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UNIVERDILY OF 
ILLINOIS LIBRARY 
AT URBANA-CHAMPAIGN 
MATHEMATICS 





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TREATISE 


ON 


oa G EB ay A, 


BY ELIAS LOOMIS, AM., 


PROFESSOR OF MATHEMATICS AND NATURAL PHILOSOPHY IN THE UNIVERSITY OF THE 
CITY OF NEW YORK, MEMBER OF THE AMERICAN PHILOSOPHICAL SOCIETY, 
OF THE AMERICAN ACADEMY OF ARTS AND SCIENCES, AND 
AUTHOR OF “ELEMENTS OF GEOMETRY. 


SECOND EDITION. 


NEW YORK: 


HARPER & BROTHERS, PUBLISHERS, 
82 CLIFF STREET. 


1847. 





TOV RS gr 
AE A au 







Entered, according to Act of Congress, i in the year one thousand — i a 
; eight hundred and forty-seven, by 8s 


Harrer & Broruers, meres 


ss $n the Clerk’s Office of the District Court of the Southern gs 
of New York. 





MATHEMATICS Lonspy 





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PREFACE, 


Tue present Treatise is designed to supply a deficiency 
which has long been felt. Having been engaged for many 
years in imparting instruction in Algebra, I have found no 
text-book entirely adapted to the wants of my pupils. Some 
treatises, from their extreme conciseness, and their aiming 
at the utmost rigor of demonstration, are too difficult for the 
majority of students, while others are far behind the present 
state of science. The present Treatise was specially de- 
signed for the use of the students of New York University, 
but it is believed to be adapted to the wants of students gen- 
erally in American colleges. It is designed for youth of fif- 
teen or sixteen years of age, who are supposed to possess 
ordinary abilities and aptitude for study. I have not, there- 
fore, attempted to demonstrate every principle in the most 
general and rigorous manner. Such demonstrations would 
often be unintelligible and repulsive to the majority of students. 
Nor, on the contrary, have I contented myself with mere me- 
chanical rules. Ihave aimed to follow the natural order of the 
youthful mind, deducing general principles from particular ex- 
amples, subjoining afterward a complete demonstration, when- 
ever it was thought that its force would be clearly apprehended. 
I have aimed to lead the student to generalize every principle. 
For this purpose, nearly every problem is twice stated ; first 
in a restricted form, and afterward in a more general one; 
and I have dwelt upon the interpretation of some peculiar 


vi PREFACE. 


results obtained in the discussion of general problems. ‘It is 
hoped the student will thus be led to perceive that the solution 
of particular problems is not the most important province of 
Algebra. | 

I have bestowed considerable pains upon the general theory 
of Equations. Each Proposition is distinctly enunciated, and 
illustrated by appropriate examples. Itis believed that Sturm’s 
Theorem is here exhibited in so simple a form, that youth of 
ordinary abilities may learn to apply it with facility and pleas- 
ure. The admirable method of Horner has been explained at 
length, and numerous examples are given of its application. 
The entire work was composed, not exclusively for the best 
scholars, nor for such as are unwilling to study, but rather for 
the majority of our students; and as such, it is commended to 
the attention of teachers of Mathematics. 


CONTENTS. 


SECTION I. 
DEFINITIONS AND NOTATION. 


DEFINITIonSs.—Difference between Algebra and Arithmetic ..0c-.-ceceeeecce-ee 
Premiere AGCItION, 800 SUDEACLION 4 cen mnenne wpe nsenessecews dasehenaence ouees ad 
Pe RA OUCATION ONG LIIVISION «acca bo neess necawcose seuss saedshdeekeceesnu 
Coefficient —-Exponent.—Power.— Root -......00-eee cece cece ssnncecccsssseees 
Algebraic Quantity —Monomial.—Polynomial . ........---.--2--0+ ti, Sisdaetes P 
Degree of a Term.—Homogeneous Polynomials.......-. sebeowencwe Tebveca sews 7 


SECTION II. 


ADDITION «2 ccccccncccnaccacceucceuccasacee 


SECTION III. 


SUBTRACTION eancace Seeeceseeceaecenuaecaeseseana 


SECTION IV. 
MULTIPLICATION. 


Case of Monomials.—Rule for the Exponents .........-ccccncncevcccccccccccce 
Case of Polynomials.—Rule for the Signs . 2... ..22 20.2 scc cece eee ee cece ewes eens 
Degree of a Product.—Number of Terms in a Product...........---------+.---- 
Theorems proved.—Quantities resoived into Factors. -...-.....---0+e-eeeee--e- 
Multiplication by detached Coefficients .........0...essecccceeee red atte cy ogee 


SECTION V. 
DIVISION. 


Case of Monomials.—Rule of Exponents............ Ceurenddewcaedan cbvatewedn 
Negative Exponents.—Symbol @ .. 2... cscs sn cece coeencees Sadie sdadde Si sle da 
EO NOMUAIS Ss sy ce cccesvenac apse ndsnecanehedewshivdecabepheuredoteesé 
SG a igs ak cp laobedewe vetke dur ed bln dhe bveels bcues 
Repeoume rOSOLVGG INtO WECCOTE wasieuedecienacewskerssuceerwddveaed Puan danced 
bavision Uwaetached Coofiicionts. 56s o.s.ssccsveneedverssaseadevevaves cei 


SECTION VI. 
FRACTIONS. 


+ Fundamental Principles.—Signs of the Terms . ....-+...--2.-s0 sce cccccescnee 2 
Tereduce.a Fraction. to lower Perm, ... ic do edsee. svecdcsvesseccubee Sveceeson 
To reduce a Fraction to an entire or mixed Quantity ..........-.s00. Sdeesecses 


12 


16 


Vill CONTENTS. 

, : ; Page 
To reduce a mixed Quantity to the Form of a Fraction ........-sseceee vompebas 47 
To reduce Fractions to a common Denominator.....-.----.++: Wee cs henna sca. «648 
Agaition end Sabtraction of Fractions .....5-ccccecuccccssccempeacnavcpain we. 00 
Moulplication and Division of Fractions ....2...0.0ssss+cacaieees ssccesamne 52 

SECTION VII. 
SIMPLE EQUATIONS. 
Definitions —Axioms employed vw su - pee opie ce wes sy ccdipecewecencccccsconces 58 
Equations solved by Subtraction and Addition. .........-c0--eseeee cece seen enee 60 
Equations solved by Division and Multiplication ........-..--2------eee-----ee- 61 
Equations cleared:of Wraction8....0..-ccc cope cuvecean os oshe¥ael cen peeeerineee 62 
Solntionof Problems .. 2.22 see nvcnsneedliuasmansee} veh ub euis seen Cee 65 
SECTION VIII. 

EQUATIONS WITH TWO OR MORE UNKNOWN QUANTITIES. 
BHlimination by 6Substitosion....<sn<s.00590 00 au hd Omen hwss ban Suuteeeee cored 78 
By Comparison.—By Addition and Subtraction 2... .-.....0eese seen ewe cecennne 79 
Equations containing three unknown Quantities .........ececeeeee ee een serene 86 
Equations containing m unknown Quantities.......2.... sc ce cece ne ewww cee ecee &8 

SECTION IX. 
DISCUSSION OF EQUATIONS OF THE FIRST DEGREE. 

Positive Values of z—Negative Values .. 5... 2. cc esce wee smncbenanac sk ennpeee 92 
Infinite. Values.-Indeterminate.-V ales? U3 aS AS «nc co cae noc cae ee cava s cee 96 
Zero and Infinity ..... -ecccececccccccccccessccvennas cecccegwovegecccccsssss 98 
Inequalities 2... scence cccccveeneduvewwastwenwa cece cccscops sib ceseescmecus 100 


SECTION X. 
INVOLUTION AND POWERS. 


Involution of Monomials.—Sign of the Result ....-...-- ce eceec concen ccceeeee-- 105 
Involution of Polynomials ........ Bf RI A Epo Sa Se pe 108 


SECTION XI. 


EVOLUTION AND RADICAL QUANTITIES. 


To extract. a-Root-of a Monomial: 2s .2s3s8052 5067. code cede Jestcccecvadscouawes 110 
Sign of the Root.—Square Root of a Trinomial -.......--.. 202s cece cece eee eeee 112 
Irrational Quantities —Fractional Exponents ..........--- 22 - cece cece ene e eens 115 
To reduce Surds to their most simple Forms -..... 2... -200-2-- cece cece ene n eens 117 
To reduce a Rational Quantity to the Form of a Surd.........-..-2--eeee ences 120 
To reduce Surds te-a-common-Index : 225525525555 bsce cee dbdd cece eeessu women 121 
Addition and Subtraction of Surd Quantities 7.2. .-0.-- cece ec ote were nee 122 
Multiplication and Division of Surd Quantities ....--..---.-------- eee eee e eens 124 
Involution and Evolution of Surd Quantifies .... 250. -.050 00s. 5-05 snceeneeeeee 127 
To find Multipliers which shall render Surds Rational .............-.----------- 129 
To reduce a Surd Fraction to a Rational Numerator or Denominator .....--..... 131 
To free an Equation from Radical Quantities......-...-.-.-sseenee- bade J Seats 133 


Calculus of Imaginary Expressions .........20.--sseeeeees ita deletes cavopece 135 


card 


CONTENTS. 1X 
Ni SECTION XII. 
EQUATIONS OF THE SECOND DEGREE. 

é Page 
Rem UTG QLUSCPAMEME ioe ccc. coco c ce cf ewes snconcssvucesu Ge dunctcavestenlOt 
meramumrrte COnmnlete. CQLORGVELIGN obs te fom oo at bo cect ceesescédensesnut sakacaas 142 
Solution of the Equation 22 pa meg. 2. cence disicinc ccc e cone ccc cdae sce apecicccs 148 
Quadratic Equations containing two unknown Quantities ........-...-.-..--.-- 156 
Discussion of the general Equation of the Second Degree........ adie Wi i sn gs 161 
Discussion of particular Problems. ...............-..-- ce cweNe lat Be OPE Lis: 
Double Values of z—Imaginary Values ........-....--.-220- E'S a Re © - 169 

SECTION XIII. 

RATIO AND PROPORTION. 

Definitions.—Ratios compared with each other .....-....-.--..-- Sey ROC eee 171 
Proportion.—Product of Extremes and Means ..........- Votes ee ees e odes 175 
Equal Ratios.—Alternation.—Inversion .-... 2.22 ec cca cece ccc cen ence cece nnen 178 
Composition.—Division.— Conversion .....-...--.-+-+2+-+----- So BE eR 179 
Like Powers of Proportionals.— Continued Proportion ............--------+.--- - 180 
SERIO EL, F TOPOLLION, —— V ALIALION Galatia s veccwaveccvsnegeenvendaccenccsen 182 

SECTION XIV. 

PROGRESSIONS. 

Arithmetical Progression—Last Term.—Sum of the Terms: ............----- ~as"l68 
Geometrical Progression—Last Term.—Sum of the Terms .............-.------ 194 
Progressions having an infinite Number of Terms..........-.....------e-eeeee- 198 
Srarmonical Propression....-.-«recsersedeurte trodes Chee BR I ‘ 200 

SECTION XV. 

GREATEST COMMON DIVISOR.—PERMUTATIONS AND COMBINATIONS. 

coment Common. Divisor.—sHOWeMGund). «<0 s «das ow tice ese dee ccsccccvcecseses 202 
Rule applied to Polynomials .............--..- ten Pe aa ou shuts ware ante 204 
Permutations, and Combinations ..........csceecesccceccncce bide Pena eee ae - 205 

SECTION XVI. 

INVOLUTION OF BINOMIALS. 

DECI GES UMMSER Ls yes Va ao seca ec Conk GE Matin wiles aes ciao densunade 210 
Pema cie te xponents.—-Coofiicionts . 22... ..ccseacesuconescsesuserscéacses cut 211 
eM CONTE oc koa wale ale Ok o HEMT TN lc cavinlaahivae ce adamecukans tae - 213 
Geer aponod to any Polynomial .... .. 2. 1-04 <ps>-- cadens hoes ce wewasess 215 
(VEX TOLENG 16 NOGAUIVE acs tcc ne aceesecdadntsiwenccescenecausecds 217 
Wrehenthe Exponent is a Kraction. >. .....s0scagserbis¥abidatie vvinccvcevecsa 218 
When the Exponent is a Negative Fraction ....-... 25.22.22 +s ee ceececeecenees 219 
Theorem applied to find the Roots of Numbers ........-..---....see00- wrens 220 

SECTION XVII. 

EVOLUTION OF POLYNOMIALS. 

Method of extracting the Square Root .........-.---.-02--eeeeeee sawpente wacwcw eee 


To extract the Cube Root of a Polynomial ............nsscccccccccces eens vc R84 


x CONTENTS. 


Page 
To extract any Root of a Polynomial ......... Pl he oo cab sabes ss ceuacanapaeeses 227 
Square Root of at-./b6.—Formaula.. ... 2... ddnc cone be cosn sues duces cu een 2 228 
SECTION XVIII. 

INFINITE SERIES. 
Definitions.—Orders ofa itlerencess — 5 000i. nase los a «owes an eee saa ea ene 233 
To find the nth Term pr aMseries .'.... sence «00+ => op ee ie eae aan ee 235 
To find ‘the Bum of Terms of a Series ©....-..-.2--» «+.» ve neee sic haere 236 
To find any Term of a Series by Interpolation -...........cebebeeacceeesens naan 238 
Fractions expanded into Infinite:Series:.- 7.22... .......0cccemennenas soee eee 239 
Infinite Series obtained by extracting the Square Root «sone catgie sc oak Cena 240 
Method of unknown Coefficients.—Rule..... 2.2.2.0 .ececceccenccccctccccccccs 241 

SECTION XIX. 

GENERAL THEORY OF EQUATIONS. 

Definitions —General Form of Equations ..-.... ween cee cencwsnemverassceens 246 
An Equation whose Root is a is divisible by r—@ ....-.22 2+. - eee eee ween wees 247 
An Equation of the mth Degree has m Roots.......--..2ccccccencccceccccccnce 248 
Law of the Coefficients.of every Equation. 2<. d,s. .n2-«.-~ 00> «sla sean eee 251 
What Equations have no Fractional Roots ........ Penn's an decane eh Rete 253 
How the Signs of all the Roots may be changed.....-........--2ecceeenncneeee 254 
The Number of imaginary Roots must be even-.......--..-.-cesseecs emeneeeen 254 
Number of Positive and Negative Roots <..0)-- swash. ss nyseweweiene ee ee 255 
The Roots may be increased or diminished by any Quantity.......-..-..-...--- 257 
The second Term of an Equation taken away . ~-..---- cece eee e eee cccececceces 258 
Dhe Limits, of a Root determined ein ..0ic,as0.s,<nmieceins,minimnsm/nccanninen pislaistele mie ake 259 
Derived Polynomials,—E qual Roots .....00=csenccnccusnieucie«s «> sons tneeee 260 
‘Thoorem of Sturm .. 0. o.-c-vescccenncancuss pas oon woieinih «elena: ikke aan 263 

SECTION XX. 

SOLUTION OF NUMERICAL EQUATIONS, 

To find the Integral Roots of an Equation. .........0...0cc0cccceccs cccesecce sens: 209 
Horner’s Method for Incommensurable Roots.......------ eee ec cecenn cee nccccs 272 
Eiquations of the fourth and higher Degrees ......-.. 2.55. tccs eee mesa seese 282 
Newton's Method of Approximation 22. £2 2r.0 Jin ee coe conn cn cnnep Saedeeesi ane 286 
Method of Double Position. «cop pa meicese <nccmmemennreey sed scons wcescuseege eee 289 
fluc ditterent Roots of Unity | can sic cies ewesewresovcecescttvsuesauuaey aan 291 

SECTION XXI. 

LOGARITHMS. 

Tiozarithms.—Hale for the Index. . ..0..n00-n-00cmwaminie= wit He decmhs nl owe canbe 294 
Multiplication and Division—Involution and Evolution. ........-.-..--.-------- 296 
Computation of Logarithms «2 20 .. 2- ea semkiomntla dale lominate~ Gots siniala wis ciate mia 301 
DiOGEPAUNS OUNCE as acer boa cp an snd i= vwaben sees neues s eb adseecccess aun 303 
Naperian Logarithms.—Common ieadthns ewe e iene Cees wcceccce sass pe 306 
Exponential Equations solved. . - sin feoe So cee ewwee nce ccc cence ccccasseneeneeees 310 
Compoun Interest.—Increase of Population ....c.0ccs0c cece eceseccccecccccece 312 


ALGEBRA. 


SECTION I. 
PRELIMINARY DEFINITIONS AND NOTATION. - 


(Article 1.) Wide neth is capable of increase or diminution, 
or will admit of mensuration, is called magnitude or quantity Y. 

_A sum of money, therefore, is a quantity, since we. may in- 
erease it or diminish it. A line, a surface, a weight, and other 
things of this nature, are quantities ; but an 7dea is not a quantity. 

(2.) Mathematics is the science of quantit y, or the science 
which investigates the means of measuring quantity. The 
operations of the. mind, therefore, such as memory, imagina- 
tion, judgment, &c., are not subjects of mathematical investi- 
» gation, since they are not quantities. 

(3.) Mathematics is divided into pure and mized. Pure 
mathematics comprehends all inquiries into the relations of 
magnitude in the abstract, and without reference to material 
bodies. It embraces numerous subdivisions, such as Arith- 
metic, Algebra, Geometry, &c. ? 

In the mixed mathematics these abstract principles are ap- 
plied to various questions which occur in nature. Thus, in 
Surveying, the abstract principles of Geometry are applied to 
the measurement of land; in Navigation, the same principles 
are applied to the determination of a ship’s place at sea; in 
Optics, they are employed to investigate the properties of, 
light; and in Astronomy, to determine the distances of the 
heavenly bodies. 

(4.) Algebra is that branch of mathematics which enables us, 
by means of letters and other symbols, to abridge and generalize 
the reasoning employed in the solution of all questions relating 
to numbers. . 

A 


4 PRELIMINARY DEFINITIONS AND NOTATION. 


(10.) The solution of a problem is the’ process by which we 
obtain the answer to it. A numerical solution is the obtaining 
an answer in numbers. <A geometrical solution is the obtaining 
_an answer by the principles of geometry. A mechanical so- 
lution is one which is gained by trials. 

(11.) The principal symbols employed in Algebra are the 
following : 

The sign + (an erect cross) is named plus, and is employed 
to denote the addition of two or more numbers. Thus, 5+3 
signifies that we must add 3 to the number 5, in which case 
the result is 8. In the same manner, 11+6 is equal to 17; 
14+-10 is equal to 24, &c. | 

We also make use of the same sign to connect several num- 
bers together. Thus, 7+5+9 signifies that to the number 7 
we must add 5 and also 9, which make 21. 

So, also, the sum of 8+5+13+11+1+3-+10 is equal to 51. 

(12.) In order to generalize numbers we represent them by 
letters, as a, b,c, d, &c. Thus the expression a+b signifies 
the sum of two numbers, which we represent by @ and 0, and 
these may be any numbers whatever. In the same manner, 
m+n+p+x signifies the sum of the numbers represented by 
these four letters. If we knew, therefore, the numbers repre- 
sented by the letters, we could easily find by arithmetic the 
value of such expressions. 

The first letters of the alphabet are commonly used to rep- 
resent known quantities,.and the last letters those which are 
unknown. 

(13.) The sign — (a horizontal. line) is called minus, sa in- 
dicates that one quantity is to be subtracted from another. 
Thus, 8—5 signifies that the number 5 is to be taken from the 
number 8, which leaves a remainder. of 3. In like manner, 
12—7 is equal to 5, and 20—14 is equal to 6, &e. 

Sometimes we may have several numbers to subtract from 
a single one. Thus, 16—5—4 signifies that 5 is to be subtract- 
ed from 16, and this remainder is to be further diminished by 
4, leaving 7 for the result. In the same manner, 50—1—38—5 
—7—9 is equal to 25. So, also, a—b signifies that the number 
designated by a is to be diminished ce the number designated 
by 6. 

Quantities preceded by the sign + are called positive quan- 


PRELIMINARY DEFINITIONS AND NOTATION. 5 


tities; those preceded by the sign —, negative quantities. 
When no sign is prefixed to a quantity, + is to be understood. 
Thus, a+b—c is the same as +a+b—c. 

(14.) The sign X (an inclined cross) is employed to denote 
the multiplication of two or more numbers. Thus, 3X5 signi- 
fies that 3 is to be multiplied by 5, making 15. In like man- 
ner, a@Xb signifies a multiplied by b; and axbxXc signifies the 
continued product of the numbers designated by a, b, and c, 
and so on for any number of quantities. 

Multiplication is also frequently indicated by placing a point 
between the successive letters. Thus,a.b.c.d signifies the 
same thing as axbxcxd. S 

Generally, however, when numbers are represented by let- 
ters, their multiplication is indicated by writing them in suc- 
cession without the interposition of any sign. Thus, a b sig- 
nifies the same thing as a.b or aXb; and a bc d is equivalent 
to a.b.c.d, or aXbxcxXd. 

It must be remarked that the notation a@.b or a b is seldom 
employed except when the numbers are designated by letters. 
If, for example, we attempt to represent the product of the 
numbers 5 and 6 in this manner, 5.6 might be confounded 
with an integer followed by a decimal fraction; and 56 would 
be read fifty-six, according to the common system of nota- 
tion. 

The multiplication of numbers may, however, be expressed 
by placing a point between them, in cases where no ambiguity 
can arise from the use of this symbol. Thus, 1.2.3.4 is 
sometimes used to represent the continued product of the num- 
bers 1, 2, 3, 4. | 

(15.) When two or more quantities are multiplied together, 
each of them is called a factor: Thus, in the expression 7X5, 
7 is a factor, and so is 5. In the product abc there are three 
factors, a, b, c. 

When a quantity is represented by a letter, it is called a 
literal factor, to distinguish it from a numerical factor, which 
is represented by an Arabic numeral. Thus, in the expression 
5ab, 5 is a numerical factor, while a and b are literal factors. 

(16.) The character ~ (a horizontal line with a point above 
and below) shows that the quantity which precedes it is to be 
divided by that which follows. 


6 PRELIMINARY DEFINITIONS AND NOTATION. 


Thus, 246 signifies that 24 is to be divided by 6, making 4. 
So, also, a~b is a divided by b. 

Generally, however, the division of two numbers is indi- 
cated by writing the dividend above the divisor, and drawing 
a line between them. 

Thus, 24+6 and a~b are usually written = and $. 

(17.) The sign = (two horizontal lines) when placed be- 
tween two quantities, denotes that they are equal to each 
other. 

Thus, 7+6=183 signifies that the sum of 7 and 6 is equal to 
13. So, also, $1=100 cents, is read one dollar equals one 
hundred cents; 3 shillings=86 pence, is read three shillings 
are equal to thirty-six pence. In like manner, a=b siggaiibs 
that a is equal to b; and a+b=c—d signifies that the sum Of 
the numbers designated by a and b is equal to the difference 
of the numbers designated by c and d. 

(18.) The symbol > is called the sign of inequality, and 
when placed between two numbers, denotes that one of them 
is greater than the other, the opening of the sign being turned 
toward the greater number. 

Thus, 8<5 signifies that 3 is less than 5, and 11>6 denotes 
that 11 is greater than 6. So, also, a>b shows that a@ is 
greater than b, and c<d shows that c is less than d. 

(19.) The coefficient of a quantity is the number or letter 
prefixed to it, showing how often the quantity is to be taken. 

Thus, instead of writing a+a+a+a-+a, which represents 
five a’s added together, we write 5a, where 5 is the coefficient 
of a. In like manner, 10ab signifies ten times the product of 
aand b. The coefficient may be either a whole number or a 
fraction. ‘Thus, 2a signifies three fourths of a. When no co- 
efficient is expr bent 1 is always to be understood. Thus, la 
and a signify the same thing. 

The coefficient may be a letter as well as a Sota In the 
expression mx, m may be considered as the coefficient of z, 
because z is to be taken as many times as there are units in m, 
If m stands for 5, then mz is 5 times zx. 

In 4axz, 4 may be considered as the coefficient of az, or 4a 
may be considered as the coefficient of z. 

(20.) The products formed by the successive multiplication of 
the same number by itself are called the powers of that number. 


PRELIMINARY DEFINITIONS AND NOTATION. 7 


Thus, 2x2=4, the second power of 2. 
2X2x2=8, the third power. 
2X2x2x2=16, the fourth power, &c. 

So, also, 3X 3=9, the second power of 3. 

3X 3X 3=27, the third power, &c. 

Also, aX a=aa, the second power of a. 
aXaxXa=aaa, the third power, &c. 

In general, any power of a quantity is designated by the 
number of factors which form the product. 

(21.) For the sake of brevity, powers are usually expressed 
by writing the root once, with a number above it at the right 
hand, showing how many times the root is taken as a factor. 
This number is called the exponent of the power. 

Thus, instead of 

aa, We write a’, where 2 is the exponent of the power. 

aaa, “  a*, where 3 is the exponent of the power. 
aaaa, “ a’, where 4 is the exponent of the power. 

aaaaa, “ a’, where 5 is the exponent of the power, &c. 

When no exponent is expressed, 1 is always understood. 
Thus, a‘ and a signify the same thing. 

Exponents may be attached to figures as well as letters. 
Thus, the product of 3 by 3 may be written 3°, which equals 9 


3 3x3x3 4 3 “ 27 
66 8x3x38x38 66 3", 66 8] 
- 3X3X3X3X3 a oy " 2438 


(22.) A root of a quantity is a factor, which, multiplied by 
itself a certain number of times, will produce the given quan- 
tity. 

The symbol Vis called the radical sign, and when pre- 
fixed to a quantity denotes that its root is to be extracted. 
Thus, 

¥9, or simply 9, denotes the square root of 9, which is 3. 

Y 64 denotes the cube root of 64, which is 4. 

4/16 denotes the fourth root of 16, which is 2. 

So, also, 

a, or simply ,/a, is the square root of a. 

a denotes the third or cube root of a. 

Ya denotes the fourth root of a. 

a denotes the nth root of a, where m may represent any 
number whatever. 





8 PRELIMINARY DEFINITIONS AND NOTATION. 


The number placed over the radical sign is called the index 
of the root. Thus, 2 is the index of the square root, 3 of the 
cube root, 4 of the fourth root, and n of the nth root. The in- 
dex of the square root is usually omitted. Thus, instead of 
Yab, we usually write /ab. 

(23.) When four quantities are proportional, the proportion 
is expressed by points, as in arithmetic. Thus, a:b::¢:d 
signifies that a has to 6 the same ratio which c has to d. 

(24.) A vinculum , or a parenthesis (_ ), indicates that 
several quantities are to be subjected to the same operation. 

Thus, a+b-+cxd, or (a+b+c) Xd, denotes that the sum of 
a, b, and c is to be multiplied by d. But a+b-+cxd, denotes 
that c only is to be multiplied by d. 

When the parenthesis is used, thé sign of rnuleittaneeel is 
generally omitted. Thus, (a+b+e)xd is the same as (a+b 
+ce)d, or d(a+b+c). 





(25.) Three dots *.* are sometimes employed to denote - 


therefore or consequently. 

A few other symbols are employed in algebra, in addition to 
those here enumerated, which will be explained as they occur. 

(26.) Every number written in algebraic language, that is, 
by aid of algebraic symbols, is called an algebraic quantity, or 
an algebraic expression. 

Thus, 3a is the algebraic expression for three times the 
number a. 

4a’ is the algebraic ll for four times the square of 
the number a. 

Zab‘ is the algebraic expression for seven times the third 
power of a multiplied by the fourth power of b. 

(27.) An algebraic quantity, not composed of parts which 
are separated from each other by the sign of addition or sub- 
traction, is called a monomial, or a quantity of one term, or 
simply a term. 

Thus, 2a, 5bc, and Txy* are eee te 

(28.) An algebraic expression, which is composed of sever al 
terms, is ditled a polynomial. 

Thus, a+2b+5c—d is a polynomial. 

A polynomial consisting of two terms only, is usually called 
a binomial; one consisting of three terms is called a trinomial. 

Thus, 8a+-5b is a binomial, and a+38bc+<ay is a trinomial. 


PRELIMINARY DEFINITIONS AND NOTATION. 9 


(29.) The numerical value of an algebraic expression is the 
result obtained when we attribute particular values to the 
letters. 

Suppose the expression is 2a°d. 

If we make a=2 and b=3, the value of this expression will 
be 2X2X2x3=24. 

If we make a=4 and b=3, the value of the same expression 
will be 2X4X4xX3=96. 

The numerical value of a polynomial is not affected by 
changing the order of the terms, provided. we preserve their 
‘respective signs. 

The expressions a?+2ab+b’, a?+b'?+2ab, b’+2ab+a’, have 
all the same numerical value. 

Thus, if a=5 and b=2, the value of a’ will be 25, that of 
2ab will be 20, and b’ will be 4; and if these numbers are 
added together, their sum will be the same in whatever order 
they are placed. Thus, 


25 25 20 ROSE 4 4'y 
20 4 25 ned 25 20 
4 20 4 25 20 25 
49 49 49 49 49. . 49 


- (80.) Each of the literal factors which compose a term is 
called a dimension of this term; and the degree of a term is 
the number. of these factors or dimensions. A numerical co- 
efficient is not counted as a dimension. 

Thus, 3a is a term of one dimension, or of the first degree. 

5ab is a term of two dimensions, or of the second degree. 

6a*bc* is a term of six dimensions, or of the sixth degree. 

In general, the degree, or the number of dimensions of a 
term, is equal to the sum of the exponents of the letters con- 
tained in the term. 

Thus, the degree of the term 5ab’cd* is 1+2+1+8 or 7; 
that is, this term is of the seventh degree. 

(31.) A polynomial is said to be homogeneous when all its 
terms are of the same degree. 

Thus, 2a—38b-+c, is of the first degree and homogeneous. 
3a°—4ab+b’,.is of the second degree and homogeneous. 
2a°+3a’c—4c'd, is of the third degree and homogene- 

ous. 
5a’—2ab+c, is not homogeneous. 


10 PRELIMINARY DEFINITIONS AND NOTATION. 


(32.) Like or similar terms are terms composed of the same 
letiers affected with the same exponents. 

Thus, 3ab and 7ab are similar terms. 

5a°c and 8a’c are also similar terms. 

But 3ab? and 4a’) are not similar ; for, although they contain 
the same letters, the same letters are not affected with the 
same exponents. 

(33.) The reciprocal of a quantity is the quotient arising 
from dividing a unit by that quantity. 

Thus, the reciprocal of 2 is 1; the reciprocal of a is + 

(34.) A few examples are here subjoined, to exercise the. 
learner on the preceding definitions and remarks. 

Examples in which words are to be converted into algebraic 
symbols. 

Ex. 1. What is the algebraic expression for the following 
statement? The second power of a, increased by twice the 
product of a and b, diminished by c, and increased by d, is 
equal to seventeen times f. 

Ans. a’+-2ab—c+d=17f. 

Hix. 2. The quotient of three divided by the sum of x and 
four, is equal to twice b diminished by eight. 

Ex. 3. One third of the difference between six times x and 
four, is equal to the quotient of five divided by the sum of a 
and b. 

Kx. 4. Three quarters of x increased by five, is equal to 
three sevenths of 6 diminished by seventeen. 

dix. 5. One ninth of the sum of six times z and five, added 
to one third of the sum of twice z and four, is equal to the 
product of a, b, and c. 

Ex. 6. The quotient ar ising from dividing the sum of a and 
b by the product of ¢ and d, is equal to four times the sum of e, 
fig, and h, , 

(35.) Haxamples in which the algebraic signs are to be trans- 
lated into common language. 

PGs eat, wh te 
bc ath 

Ans. The quotient arising from dividing the sum of z and a 
by 0b, increased by the quotient of 2 divided by c, is equal to 
the quotient of d divided by the sum of a and b. 

Ez. 2. Ta’+(b—c) X (dt+e)=g+h. 


: 


PRELIMINARY DEFINITIONS AND NOTATION. 11 


How should the preceding example be read, when the first 
parenthesis is omitted ? 
atg ,6—4m_h 


ies oe a 
sacha 2h+c 
Ex. 4. 4V ab—25=5 


Ex. 5. 2aVb'—ac=5(h+d+z2). 
Ex. 6. ae ey 
(36.) Find the value of the following expressions, when a= 
6, b=5, and c=4. 
Ex. 1. a@+8ab—c’. 
Ans. 36+90—16=110. 


Hix. 2. ax (a+b)—2abe. 





Ans. 156. 
a’ ; 
Ez. 3 mal Bales 
Ans. 28. 
Ez. 4. c+ soit 
V 2act+c? 


Ex. 5. /b?—ac+V2ac+c’. 
Ex. 6. 3/c+2av 2a+b+2c. 
Ex. 7. (8/c+2a) V2a+b+2c¢. 


axv*+b? 
he Bape Er tanks i an Seay ne Se 
xr—a’—c 


and z=6; what is its numerical value? 


Hx. 8. In the expression 


SECTION II. 


ADDITION. 


(37.) Addition is the connecting of quantities together by 
means of their proper signs, and incorporating such as can be 
united into.one sum. 

It is convenient to distinguish three erm 


CASH 


When the quantities are similar and have the same signs. 


RULE. 


Add the coefficients of the several quantities together, and to 
* their sum annex the common letter or letters, prefixing tle com- 
mon sign. 

Thus, the sum of 8a and 5a is obviously 8a. So, also, —8a 
and —5a make —8qa; for the minus sign before each of the 
terms shows that they are to be subtracted, not from each 
other, but from some quantity which is not here expressed ; 
and if 3a and 5a are to be successively subtracted from the 
same quantity, it is the same as subtracting at once 8a. 


EXAMPLES. 
38a = —8ab = 2+ Bx a—2x* Qat+ y? 
5a —6ab 5b+- Tx 4a—3x° 5a+2y* 
7a — ab b+2x 3a— 5x°* 9a+3y? 
a —Tab 4b+3z hy fee 4a+6y’ 
16a. —l%ab. /gb+isx arnt 0.77 


The learner must continually bear in mind the remark of 
Art. 13, that when no sign is prefixed to a quantity, plus is al- 
ways to be understood. 


ADDITION. La 


CASE II. 


(38.) When the quantities are similar, but have different 
signs. 


sy 


RULE. 


Add all the positive coefficients together, and also all those 
that are negative; subtract the least of these results from the 
greater; to the difference annex the common tees or letters, and 
prefix the sign of the greater sum. 

Thus, eieda of 7a—4a, we may write 3a, § since these two 
expressions obviously have the same value. 

Also, if we have 5a—2a+38a—a, this signifies that from 5a 
we are to subtract 2a, add 8a to the remainder, and then sub- 
tract a from this last sum, the result of which operation is 5a. 
But it is generally most convenient to take the sum of the pos- 
itive quantities, which in this case is 8a; then take the sum of 
the negative quantities, which in this case is 83a; and we have 
8a—8a or 5a, the same result as before. 


EXAMPLES. ok 
=3a 6z+5ay 2ay— T —2a°x —6a’+2b 
+7a —8x+2ay — ay+ 8 ax 2a°— 3b 
+8a z—6ay -* 2ay— 9° -+~3a°'x = —5a'—8) 
o—,G 22+ ay 3ay—11 Tax 4a’°—2b 








CASE IIL 


(39.) When some of the quantities are dissimilar. 


RULE. 


Collect all the like quantities together, by taking their sums or 
differences as in the two former cases, and set down those that 
are unlike, one after the other, with their proper signs. 

Unlike quantities can not be united in one term. Thus, 2a 
and 3b neither make 5a nor 5b. Their sum can only be writ- 
ten 2a+3b. 


14 ADDITION. 


EXAMPLES. 

LLY — 2" 3x°y+2ax 2ax — 220 2x —18y 

or ty Sy a ee ary +10z 

x — xy —3y'a+3azr" 5x? —3x Qx*y+25y 
42° —3ry —B8z"'y— az 3x +100 1227*°y—zry 


62? — xy 5 apt Qazp-F HG —Syp, 8x" Sz” — 120 Tpy Fy FPR | 7» 

(40.) When several neriines are to be’ ‘added together, it 
is most convenient to write all the similar terms under each 
other, as in the following example. 


Ex. 1. Add together 
11bc+4ad—8ac-+ decd 


8ac+7bc—2ad+4mn 
2cd—B8ab+5ac+ an 
9an—2be—2ad+5cd 
These terms may be written thus: 
1lbc+4ad—8ac+5cd+an+4mn—3ab 
Tbc—2ad+8ac+2cd+9an 
—2bc—2ad+5ac+5cd 
Sum  16bc +5ac+12cd+10an+4mn—8B8ab. 
iz. 2. Add together the quantities 





7m+3n —14p 
3a +9n —llim 
5p —4m-+ 8n 


lln—2b— m 
Ans. 3a—2b—9m+31n—9p. 
Ex. 3. Add together 7 
4a°b + 3c°d—9m’'?n—6ab’ 
4m’n— ab’+5c’d +7a*b 
6m'’n— 5c°d+4mn’>—S8ab’ 
Imn'+ 6c°'d—5m'’n—6a’°b 
9c°d —10ab’—8m’?n+12a°b 
Ans. 17a°b+-18c?d—12m’n—25ab’?+11mn’'. 
Ez. 4. Add together 
3b— a—6c—115d—9f 
be=)/ — 7d. Ue 0a 
8a—2b—3ce+ BWet+l1lf 
‘ 38e—Tf+5b— 8c+9d 
17c—6b—Ta— 2d—5e 
Ans. —8a-+6c—109d+3le—10f. 


ADDITION. 15 


_ Ex. 5. Add together 
2ab*+3ac’+ 9b°x— Bhy’+-10ky 
2ab°*—32" — bax— 4ky*?—lbhy 
5ky — hy?—22ac?—102? — 4ab’ 
19ac’—8b'x+ 92° + 6hy + Qky’ 
Ans. —9hy’?+15ky—2ky’?—9hy— 42". 

(41.) It must be observed that the term addition is used in 
a more extended sense in algebra than in arithmetic. In arith- 
metic, where all quantities are regarded as positive, addition 
implies augmentation. The sum of two quantities will there- 
, fore be numerically greater than either quantity. Thus the 
” sum of 7 and 5 is 12, which is Se greater than either 
5 or 7. 

But in algebra we consider negative as well as positive 
quantities ; and by the sum of two quantities, we mean their 
ageregate, regard being paid to their signs. Thus the sum 
of +7 and —5 is +2, which is numerically less than either 7 
or 5. So, also, the sum of +a and —bis a—b. In this case, 
the algebraic sum is numerically the difference of the two 
quantities. 

This is one instance, among many, in which the same terms 
are used in a much more general sense in the higher mathe- 
matics than they are in arithmetic. 


SECTION JIT. 


SUBTRACTION. 


(42.) Subtraction is the taking of one quantity from anoth- 
er; or it is finding the difference between two qnantities or 
sets of quantities. 

Let it be required to subtract 8—8 from 15, 

Now 8—8 is equal to 5. 

And 5 subtracted from 15 leaves 10. 

The result, then, must be 10. But, to perform the operation 
on the numbers as they were given, we first subtract 8 from 
15, and obtain 7. This result is too small by 3, because the 
number 8 is larger by 3 than the number which was required 
to be subtracted. Therefore, in order to correct this result, 
‘the 3 must be added, and we have 

.15—8+8=10, as before. 
Again, let it be required to subtract c—d from a—b. It is 


plain, that if the part c were alone to be subtracted, the re- 


mainder would be 
a—b—c. 

But as the quantity actually proposed to be subtracted is 
less than c by d, too much has been taken away by d, and, 
therefore, the true remainder will be greater than a—b—c by 
d, and will hence be expressed by 

a b-rct-d, 
where the signs . of the last two terms are both contr ary to 
what they were given in the subtrahend. 

(43.) Hence we deduce the following general 


RULE. 


Conceive the signs of all the terms of the subtrahend io be 


SUBTRACTION. 17 


changed from + to —, or from — to +, and then collect the 
terms together, as in the several cases of addition. 

It is better in practice to leave the signs of the subtrahend 
unchanged, and simply conceive them to be changed ;. that is, 
treat the quantities as if the signs were changed; for, other- 
wise, when we come to revise the work to detect any error in 
the operation, we might often be in doubt as to what were the 
signs of the quantities as originally proposed. 





EXAMPLES. 

From 5a’—2b B5xryt+8x—2 10 —8x—Bry 4ax—2x*y 
Subtract 2a°+5b S8ry—8r—7 —2z+3 — zy 3azr— oaniee id 
Remainder 3a°—7b ~ +» 5 7 —Txe—Qry 


From 5a+4b—2c+7d From. llay+2y’—162’? 
Take 3a+2b+ c+5d Take — 4xy+6y’?— 182’ 
Remainder 2a+2b—3c+2d ey Ah 
From 6aby—4ry+4zz From 2°+2zy+y’ 
Take —8aby+5zz4+ 32ry Take 2°—2ry+y’ 
Remainder 9aby— xz—Txy tes 
From 3a°+ ax+22°—14a°x+19ax?— 42? +5a°2’ 
Take 2a°—4ax+ x’ —15a°x+ llax’?—15a*x?—42’ 











Subtraction may be proved as in Arithmetic, by adding the 
remainder to the subtrahend.. The sum should be equal to the 
minuend. 

(44.) The term subtraction, it will be perceived, is used in 
a more general sense in algebra than in arithmetic. © In arith- 
metic, where all quantities are regarded as positive, a number 
is always diminished by subtraction. But in algebra, the dif- 
ference between two quantities may be numerically greater 
than either. Thus, the difference between +a and —b is a+. 

The distinction between positive and negative quantities 
may be illustrated by the scale of a thermometer. The de- 
grees above zero are considered positive, and those below zero 
negative. From five degrees above zero to five degrees be- 
low zero, the numbers stand thus: 

Hoot S73, +-2,)-+1, 0, — 1). —-2,:—3, —4, — 5. 

The difference between five degrees above zero and five 
degrees below zero is ten degrees, which is numerically the 
sum of the two quantities. 


B 


18 ' SUBTRACTION. 


(45.) In practice, it is often sufficient merely to indicate the 
subtraction of a poly nomial, without actually performing the 
operation. This is done by inclosing the palynomialin in a pa- 
renthesis, and prefixing the sign —. 

Thus, 5a—3b-+4c—(8a—2b+8c) 
signifies that the entire quantity 3a—2b-+8c is to be subtracted 
from 5a—3b+4c. The subtraction is here merely indicated. 
If we actually perform the operation, the expression becomes 
5a—3b+4c—3a+2b—8e, 
or 2a— b—4e. 

(46.) According to the preceding principle, polynomials may 
be written in a variety of forms. 

Thus, a— b— c+d 
is equivalent to a—(b+ c—d), 

or to a— b—(c—d), 

or to a+ d—(b+c). 

Transformations of this sort, which consist in decomposing 
a polynomial into two parts separated from each other by the 
sign —, are of frequent use in algebra. It is recommended to 
the student to write out polynomials like the above, contain- 
ing both positive and negative terms, in all the possible modes, 
including several terms in a parenthesis. 

In the following examples, let the results all be reduced to 
their simplest form. 

Ex. 1. a+b—(2a—3b) — (5a+7) — (—138a+2b) =. 

Ex. 2. 37a—5f— (8a—2b—5c) — (6a—4b+ 8h) =. 

Ex. 3. 8a*xy— 5ba*y+licry’—9y°— (aay + 3ba*y—13cxry? + 
20y°)=. 

Ex, 4, 28ax?—16a’x’ +25a°x— 18a'— (18az?+20a*x*— 24072 
—7a')=. 

(47.) It has already been remarked, in Art. 5, that algebra 
differs from arithmetic in the use of negative quantities, and it 
is important that the beginner should obtain clear ideas of their 
nature. 

In many cases, the terms positive and negative are merely 
relative. They indicate some sort of opposition between two 
classes of quantities, such that if one class should be added, the 
other ought to be subtracted. Thus, if a ship sails alternately 
northward and southward, and the motion in one direction is 


SUBTRACTION. 19 


ealled positive, the motion in the opposite direction should be 
considered negative. 

Suppose a ship, setting out from the equator, sails north- 
ward 50 miles, then southward 27 miles, then northward 15 
miles, then southward again 22 miles, and we wish to deter- 
mine the last position of the ship. If we call the northerly 
motion +, the whole may be expressed algebraically thus: 

+50—27+ 1522, 
which reduces to +16. The positive sign of the result indi- 


cates that the ship was 16 miles north of the equator. 
Suppose the same ship sails again 8 miles north, then 35 


miles south, the whole may be expressed thus: 

+50 —27+15—-224-8—35, 
which reduces to —11. The negative sign of the éeuit indi- 
cates that the ship was now 11 miles south of the equator. 

In this example we have considered the northerly motion +, 
and the southerly motion — ; but we might with equal pro- 
priety have considered the southerly motion +, and the north- 
erly motion —. It is, however, indispensable that we adhere 
to the same system throughout, and retain the proper sign of 
the result, as this sign shows whether the ship was at any time 
north or south of the equator. 

In the same manner, if we consider easterly motion +, 
westerly motion must be regarded as —, and vice versa. 
And generally, when quantities which are estimated in differ- 
ent directions enter into the same algebraic expression, those 
which are measured in one direction being treated as +, those 
which are measured in the opposite direction must be regard- 
ed as —. 

So, also, in estimating a man’s property, gains and losses 
being of an opposite character, must be affected with different 
signs. Suppose a man, with a property of 1000 dollars, loses 
300 dollars, afterward gains 100, and then loses again 400 
dollars, the whole may be expressed algebraically thus: 

+1000—300-+ 100—400, 
which reduces to +400. The + sign of the result indicates 
that he has now 400 dollars remaining in his possession. Sup- 
pose.he further gains 50 dollars and then loses 700 dollars. 
The whole may now be expressed thus: 
+1000—300 + 100—400+-50—700, 


20 SUBTRACTION. 


which reduces to —250. The — sign of the result indicates 
that his losses exceed the sum of all his gains and the property 
originally in his possession; in other words, he owes 250 dol- 
lars more than he can pay, or, in common language, he is 250 
dollars worse than nothing. 

This phraseology must not be regarded as wholly figurative ; 
for, in algebra, a negative quantity standing alone is regarded 
as less than nothing; and of two negative quantities, that 
which is numerically the greatest is considered as the least; 
for if from the same number we subtract successively num- 
bers larger and larger, the remainders must continually di- 
minish. Take any number, 5 for example, and from it subtract 
successively 1, 2, 3, 4, 5, 6, 7, 8, 9, &c., we obtain 
5—1, 5—2, 5—3, 5—4, 5—5, 5—6, 5—7, 5—8, 5—9, Ke., or 
reducing 

4,8; 21, Opel ist trop 
Whence we see that —1 should be regarded as smaller than 
nothing ; —2 less than —1; —3 less than —2,&c. 


SECTION IV. 


MULTIPLICATION. 


(48.) Multiplication is repeating the multiplicand as many 
times as there are units in the multiplier. 

When several quantities are to be multiplied together, the 
result will be the same in whatever order the multiplication is 
performed. 

This may be demonstrated in the following manner: 

Let unity be repeated five times upon a horizontal line, and 
let there be formed four such parallel lines. 


Then it is plain that the number of units in the table is equal 
to the five units of the horizontal line, repeated as many times 
as there are units in a vertical column; that is, to the product 
of 5 by 4. But this sum is also equal to the four units of a 
vertical line repeated as many times as there are units in a 
horizontal line; that is, to the product of 4 by 5. Therefore, 
the product of 5 by 4 is equal to the product of 4 by 5. For 
the same reason, 2X3xX4 is equal to 2X4X3, or 4X3XQ, or 
3X4X2, the product in each case being 24. So, also, if a, b, 
and ¢ represent any three numbers, we shall have abc equal to 
bea or cab. 

It is convenient to consider the subject of multiplication un- 
der three Cases. 


V2 MULTIPLICATION. 


CASE TL 


(49.) When both the factors are monomials. 

From Article 14, it appears that, in order to represent the 
multiplication of two monomials, such as 8abe and 5def, we 
may write these quantities in succession without interposing 
any sign, and we shall have 

sabcbdef. i 

But, according to the principle stated in the preceding ar- 

ticle, this result may be written 
3x 5abcdef, or 15abcdef. 
Hence we deduce the following 


RULE, 


Multiply the coefficients of the two terms together, and to the 
product annex all the different letters in succession. 


EXAMPLES. 


Multiply 12a 5a Tab Taxy 6xryz 

By ~~ 3b 6x 5ac 6ay ay 

Product 36ab SAC So a bank! 

From Article 48, it appears to be immaterial in what order 
the letters of a term are arranged; it is, however, generally 
most convenient to arrange them alphabetically. 

(50.) We have seen in Art. 21, that when the same letter 
appears several times as a factor in a product, this is briefly 
expressed by means of an exponent. Thus, aaa is written a’, 
the number 3 showing that a enters three times as a factor. 
Hence, if the same letters are found in two monomials. which 
are, to be multiplied together, the expression for the product 
may be abbreviated by adding the exponents of the same let+ 
ters. Thus, if we are to multiply @ by a’, we find a equiva- 
lent to aaa,and a? toaa. Therefore the product will be aaaaa, 
which may be written a°, a result which we might have ob- 
tained at once by adding together 3 and 2, the exponents of 
the common letter a. 

Hence, since every factor of both multiplier and multipli- 
cand must appear in the product, we have the following 


MULTIPLICATION. 23 


RULE FOR THE EXPONENTS. 


Powers of the same quantity may be multiplied by adding 
their exponents. 











EXAMPLES. 
Multiply 8a*be’ 2a‘b’c 5a‘b*c? 2a°b*c* 
By abcd’ 8abc* Ta‘bic'd —Ba*bc* 
Product 56a°b'c'd? /f a7 2804 3509] a, 
CASE II. 


(51.) When the multiplicand is a polynomial. 

If a+b is to be multiplied by c, this implies that the sum of 
the units in @ and b is to be repeated c times; that is, the units 
in b repeated c times must be added to the units in a repeated 
also c times. Hence we deduce the following 


RULE. 


Multiply each term of the multiplicand separately by the mul- 
tiplier, and add together the products. 


EXAMPLES. 
Multiply 83a0+2b @+2r4+1 8 3y°+5ry+2 32°+2y+2y’ 
By _4a Ax nite ealuetigees Be 
Product 12a*+8ab SLs im 

CASE IIL 


q 


(52.) When both the factors are polynomials. 

If a+b is to be multiplied by c+d, this implies that the 
quantity a+b is to be repeated as many times as there are 
units in the sum of c and d; that is, we are to multiply a+b 
by c and d successively, and add the partial products. Hence 
we deduce the following 


RULE. 


Multiply each term of the multiplicand by each term of the 
‘multiplier separately, and add together the products. 


24 MULTIPLICATION. 


EXAMPLES. 
Multiply a+b 3x +2y ax+-b 3a+. x 





By a+b 22+ 3y cet+d  2a+4e 
Product a?+2ab+b? (vi ix Ay WAL feet 

When several terms in the product are similar, it is most 
convenient to set them under each other, and then unite them 
by the rules for addition. 

(53.) The examples thus far given in multiplication have 
been confined to positive quantities, and the products have all 
been positive. We must now establish a general rule for the 
signs of the product. 

First, if +a is to be multiplied by +48, this signifies that +a 
is to be repeated as many times as there are units in 0, and 
the result is +ab. That is, a plus quantity multiplied by a 
plus quantity gives a plus result. 

Secondly, if —a is to be multiplied by +48, this signifies that 
—a is to be repeated as many times as there are units in Bb. 
Now —a taken twice is obviously —2a, taken three times is 
—3a, &c.; hence, if —a is repeated b times, it will make —ba 
or —ab. That is, a minus quantity multiplied by a plus quan- 
tity gives minus. 

Thirdly, to determine the sign of the product when the mul- 
tiplier is a minus quantity, let it be proposed to multiply 8—5 
by 6—2. By this we understand that the quantity 8—5 is to 
be repeated as many times as there are units in 6—2. If we 
multiply 8—5 by 6, we obtain 48—30; that is, we have re- 
peated 8—5 six times. But it was only required to repeat the 
multiplicand four times, or (6—2). We must therefore dimin- 
ish this product by twice (8—5), which is 16—10; and this 
subtraction is performed by changing the signs of the subtra- 
hend ; hence we have 

48—30—16+10, 
which is equal to 12. This result is obviously correct; for 
8—5 is equal to 3, and 6—2 is equal to 4; that is, it was re- 
quired to multiply 3 by 4, the result of which is 12, as found 
above. 

In order to generalize this reasoning, let it be proposed to 
multiply a—b by e—d. 

If we multiply a—b by c, we obtain ac—be. But a—b was 


MULTIPLICATION. 95 


only to be taken c—d times; therefore, in this first operation, 
we have repeated it too many times by the quantity d. Hence, 
to have the true product, we must subtract d times a—b from 
ac—bc. But d times a—b is equal to ad—bd, which, subtract- 
ed from ac—be, gives 
ac—be—ad-+bd. 

Thus we see that +a multiplied by —d gives —ad; and —b- 
multiplied by —d gives +bd. Hence a plus quantity multi- 
plied by a minus quantity gives minus, and a minus quantity - 
multiplied by a minus quantity gives plus. 

(54.) The preceding results may be briefly expressed as fol- 
lows: 

+ multiplied by +, and — multiplied by —, give +. 

++ multiplied by —,and — multiplied by +, give —. 

Or, the product of two quantities having the same sign, has 
the sign plus; the product of two quantities having different 
signs, has the sign minus. 

(55.) The whole doctrine of multiplication is therefore com- 
prehended in the foilowing 


RULE. 


Multiply each term of the multiplicand by each term of the 
multiplier, and add together all the partial products, observing 
that like signs require + in the product, and unlike signs —. 


EXAMPLE I. 
Multiply 5a'— 2a°b+ 4a’? 
By a— 4a°b+ 2b° 


Beotict —20a*b+ 8a°l?—16a'b® ° 
ees +10a'b’—4a°b'+8a°b° 
Result 5a’ —22a°b + 12a°b’— b6a'b®—4a°h +8a°d* 


5a’— 2a°b 4°’ 
Partial { i AF 








Ex. 2. Multiply 4a°—5a*b—8ab’+2b* by 2a°—3ab—40”. 
Ans. 8a°—22a*b—17a*b’?+48a’b’ + 26ab*—8d'. 
Ez. 3. Multiply 3a°—5bd+ef by —5a’+4bd—8ef. 
Ans. —15a‘+87a°bd— 29a°ef — 20b*d* +-44bdef—8e°f”. 
| Ex. 4. Multiply 2*+22°+382?+2¢+1 by 2°—2zr+1. 
> Ans. x°—22z°+1. 
Ex. 5. Multiply 44a’e—6a’bce+c* by 14a°e+6a*be—c’. 


26 MULTIPLICATION. 


Hix. 6. Saecey 3a" peeria iad! t ABN by, Bali Oo 
57D". HAS ° i As 

(56.) Since in the mnaltiplication we two vaetibadedlie every 
factor of both quantities appears in the product, it is obvious 
that the degree of the product will be equal to the sum of the 
degrees of the multiplier and multiplicand. Hence, also, if 
‘two polynomials are homogeneous, their product will be homo- 
geneous. 

Thus, in the first of the preceding examples, all the terms 
of the multiplicand being of the faunal degree, and those of. 
the multiplier of the third degree, all the terms of the product 
are of the seventh degree. For a like reason, in the second 
example, all the terms of the product-are of the fifth degree ; 
in the third example, they are of the fourth degree; and in the 
sixth example, they are of the fifth degree. 

This remark will enable us to detect any error in the mul- 
tiplication, so far as concerns the exponents. For example, if 
we find in one of the terms of a product which should be ho- 
mogeneous, the sum of the exponents equal to 6, while in all 
the other terms it is equal to 7, a mistake has evidently been 
committed in the formation of one of the terms. 

(57.) When the product arising from the multiplication of 
two polynomials does not admit of any reduction of similar 
terms, the whole number of terms in the product is equal to the 
number of terms in the multiplicand, multiplied by the number 
of terms in the multiplier. 

Thus, if we have five terms in the multiplicand and four 
terms in the multiplier, the whole number of terms in the prod- 
uct will be 5x4, or 20. In general,if there be m terms in the 
multiplicand and terms in the multiplier, the whole number 
of terms in the product will be mxXn. 

(58.) If the product contains similar terms, the number of 
terms in the product when reduced may be much less; but it 
is important to observe, that among the different terms of the 
product there are always two which can not be combined with 
any others. These are, 

1. The term arising from the multiplication of the two terms 
affected with the sneinst exponent of the same letter. 

2. The term arising from the multiplication of the two terms 
affected with the lowest exponent of the same letter. 


MULTIPLICATION. py 


For it is evident, from the rule of exponents, that these two 
partial products must involve the letter in question, the one 
with a higher, and the other with a lower exponent than any 
of the other partial products, and therefore can not be similar 
to any of them. ‘Hence the product of two polynomials can 
never contain less than two terms. 

(59.) For many purposes, it is sufficient merely to indicate 
the multiplication of two polynomials, without actually per- 
forming the operation. This is effected by inclosing the quan- 
tities in parentheses, and writing them in succession with or 
without the interposition of any sign. | 

Thus, (a+b+c) (d+e+f) signifies that the sum of a, b, and 
c is to be multiplied by the sum ef d, e, and f. 

When the multiplication is actually performed, the expres- 
sion is said to be expanded. 

(60.) The following Theorems are of such extensive appli- 
cation that they should be carefully committed to memory. 


THEOREM I. 


The square of the sum of two quantities is equal to the square 
of the first, plus twice the product of the first by the second, plus 


the square of the second. 





Thus, if we multiply a+ b 
By | a+ b 
a’+ ab 
ab+b* 
We obtain the product ai +2ab+b". 


Hence, if we wish to obtain the square of a binomial, we 
can write out the terms of the result at once according to this 
theorem without the necessity of performing an actual multi- 
plication. . 


EXAMPLES. : 
Ls (2a-+b)’=.4s 6. (5a°+Tab)’=. 
2. (a+3b)’=. | 7... (6a +6)’=. 
3. (8a+3))?=. gayi +48. (5a? +-8a7b)’?=. | 
4, (4a+3b)?.162244549°9o L+))3=. 7 - 
5. *(5a?-+b)?=.: y 10. (8+-3)2=. ed 


This theorem deserves particular attention, for one of the 


28 MULTIPLICATION. 


most common mistakes of beginners is to call the square of 
a+b equal to a’+0’. 


THEOREM II. 


(61.) The square of the difference of two quantities is equal 
to the square of the first, minus twice the product of the first and 
second, plus the square of the second. 





Thus, if we multiply a—b 
By a—b 
a— ab 
— ab+b* 
We obtain the product a —2ab+b'. 
EXAMPLES. 
1. (a— 2b)! 4) 6 (TaD). peta y ade & 
2. (2a—BbyP Hye /aultgs4. (Ta 12ab P= yPenes* 1 
3. (5a—4b)’=. , 8. “(7a"b' = 124d 
4, (6a°—2)?= 9 6at=/7ITD;) (2— 1) 
5. (6a°—82)?= (10. (4-1)'= = 


Here, also, beginners anien commit the mistake of putting 
the square of a— x; equal to a’—b’. 


THEOREM Il. 


(62.) The product of the sum and difference of two quantities 
is equal to the difference of their squares. 





Thus, if we multiply a+b 
By ab 
a’ +-ab 
* —ab—B’ 
We obtain the product a—b’, 
EXAMPLES. 
1. (2a+b) (2a—b)=. 
a (8a+4b) (8a—4b)=. 
3. (7a+z) (Ta—z)=. 
4. (Jab+z) (7ab—z)=. 
5. (8a+b) (8a—b)=. 
6. (8a+7bc) (8a—Tbc)=. 


- 


(5a°+6b*) (5a°—6b*) = 
(5z°y+3zy’) (52°y—s3xry’)=. 


® 


\ 


MULTIPLICATION. 29 


9, (8+3) (8-1)=. 
10. (4+3) (4-2)=. 


The student should be drilled upon examples like those ap- 
pended to the preceding theorems until he can produce the re- 
sults mentally with as great facility as he could read them if 
exhibited upon paper. 

The utility of these theorems will be the more apparent, the 
more complicated the expressions to which they are applied. 
Frequent examples of their application will be seen hereafter. 

(63.) The same theorems will enable us to resolve many 
complicated expressions into their factors. 


1. Resolve a’?+4ab+-40’ into its factors. 
, Ans. (a+2b) (a+2b). 
2. Resolve a?—6ab+90’ into its factors. 
3. Resolve 9a?—24ab+160’ into its factors. 
4. Resolve a*—d* into three factors. 
5. Resolve a°—D* into its factors. 
6. Resolve a*—D° into four factors. 
7. Resolve 25a*—60a’b'+360’ into its factors. 
8. Resolve n?+2n-+1 into its factors. 
9. Resolve 4m’n?’—4mn-+1 into its factors. 
10. Resolve 49a*b*— 168a°b°+ 14427)’ into its factors. 
11. Resolve n*+2n’+ 7 into three factors. 
12. Resolve 1—,', into two factors. 
13. Resolve 4—;, into two factors. 


MULTIPLICATION BY DETACHED COEFFICIENTS. 


(64.) The coefficients of a product depend simply upon the 
coefficients of the two factors, and not upon the literal parts 
of the terms. Hence we may obtain the coefficients of the 
product by multiplying the coefficients of the multiplicand sev- 
erally by the coefficients of the multiplier. To these coeffi- 
‘cients the proper letters may afterward be annexed. This 
will be best understood from a few examples. 

Thus, take the first example of Art. 52, to multiply a+b by 
a+b. 


30 MULTIPLICATION. 


The coefficients of the multiplicand are 1-++1 


e “ multiplier + 1+1 
1+1 
| 1+1 
Coefficients of the product 14+2+1 
or, supplying the letters, we obtain a’+2ab+b'*," 


which is the same result as before obtained. 
Ex. 2. Multiply 3a°+4ax2—52* by 2a°—6azr+4z’. 





Coefficients of multiplicand 3+ 4— 5 
KS multiplier 2— 6+ 4 
6+ 8—10 
—18—24+30 
' +12+16—206 
Coefficients of the product 6— 10—22+46—20 


It may seem difficult in this. case to supply the letters; but 
a little consideration will render it perfectly plain. Thus, 
3a’ X 2a’ is equal to 6a‘; hence a‘ is the proper letter to be at- 
tached to the first coefficient. For the same reason, z* is the 
proper letter to be attached to the last coefficient. Moreover, 
we see that both the proposed polynomials are homogeneous, 
and of the second degree. Hence the product must be ho- 
mogeneous, and of the fourth degree. The powers of a must 
decrease successively by unity, beginning with the first term, 
while those of z increase by unity. Hence the required pred- 
uct 1s 
6a‘ — 10a°xa—22a°x* +46ar*— 202". 

Ex. 3. Multiply z°+2°y+zy’?+y° by x—y. 

Ez. 4. Multiply 2°—32?+38zr—1 by 2’?—22+1. 

Ex. 5. Multiply 2a°—3ab’+5b° by 2a—5b. 

If we should proceed with this example precisely in the same 
manner as with the preceding, we should commit an error by 
attempting to unite terms which are dissimilar. The reason 
is, that the multiplicand does not contain the usual complete 
series of powers of a. The term containing the second power 
of ais wanting. This does not render the method inapplica- 
ble, but it is necessary to preserve dissimilar terms distinct 
from each other; and since, while we are are operating on the 
coefficients, we have‘not the advantage of the letters to indi- 
cate what are similar terms, we supply the place of the defi- 


MULTIPLICATION. 31 


cient term by a cipher. The operation will then proceed 
with entire regularity. 


2+ 0—3+ 5 

mao 5 

4+ 0—6+10 
—~10-—0+15—25 


4—10—6+25—25 
Hence the product is 
4a‘*—10a°b—6a*b’ +25ab*— 25d". 

Ez. 6. Multiply 2a°—3ab’+-5b* by 2a’— 5b’. 

Here there is a term in each polynomial to be supplied by a 
cipher. . 

The preceding examples are intended to lead the student to 
consider the properties of coefficients by themselves, and pre- 
pare him for some investigations which are to follow, particu- 
larly in Section XX. The beginner, however, in attempting 
to apply the method, must be cautious not to unite dissimilar 
terms. 


SECTION V. 


DIVISION. 


(65.) The object of division in Algebra is the same as in 
Arithmetic, viz., The product of two factors being given, and 
one of the factors, to find the other factor. 

The dividend is the product of the divisor and quotient, the 
divisor is the given factor, and the quotient is the factor re- 
quired to be found. 


CASE IL 


(66.) When the divisor and dividend are both monomials. 

Suppose we have 68 to be divided by 7. We must find such 
a factor as, multiplied by 7, will give exactly 63. We per- 
ceive that 9 is such a number, and therefore 9 is the quotient 
obtained when we divide 63 by 7. 

Also, if we have to divide ab by a,it is evident that the 
quotient will be b; for a multiplied by b gives the dividend ab. 
So, also, 12mn divided by 3m gives 4n; for 3m multiplied by 
4n makes 12mn. 

Suppose we have a* to be divided by a’. We must find a 
number which, multiplied by a’, will produce a®*. We perceive 
that a® is such a number; for, according to Art. 50, we multi- 
ply a® by a’, by adding the exponents 2 and 3, making 5. 
That is, the exponent 3 of the quotient is found by subtracting | 
2, the exponent of the divisor, from 5, the exponent of the divi- 
dend. Hence the following ' 


RULE OF EXPONENTS IN DIVISION. 
In order to divide quantities expressed by different powers 
of the same letter, subtract the exponent of the divisor from the 
exponent of the dividend. 


DIVISION. 33 


EXAMPLES. 
Divide a’? a’ CU! ht 2 y" 
By a a’ i c* h* x y” . 
Quotient a’ a Z F Peed eed at Y 344. ~ tn 


Let it be required to divide 35a° by 5a’... We must find a 
quantity which, multiplied by 5a’, will produce 35a°, Such a 
quantity is 7a°; for, according to Arts. 49 and 50, 7a’ x 5a’ is 
equal to 35a°. Therefore, 35a° divided by 5a’ gives for a 
quotient 7a°; that is, we have divided 35, the coefficient. of the. 
dividend, by 5, the coefficient of the divisor, and have sub- 
tracted the exponent of the divisor from the exponent of the 
dividend. 

(67.) Hence, for the division of monomials, we have the fol- 
lowing 


e RULE. 

1. Divide the coefficient of the dividend by the coefficient of the 
divisor. : 

2. Subtract the exponent of each letter in the divisor from the 
exponent of the same letter in the dividend. 


EXAMPLES. 
. Divide 20z° by 4z. 3 Ans: 52’. 
. Divide 25a°zy* by 5ay’. Fay 
. Divide 72ab‘z? by 120°x. A athe. 
. Divide 77a°*b'c’ by 11ab*c*. spas 
. Divide 272a°b'c’x® by 17a°b'cr" TE 
. Divide 250z’y*x' by 5ayq. a 
» Divide 48a°b’c’d by 12ab’c. ea? | 
. Divide 150a°b’cd* by 30a‘b'd’. Pied Oo es 
(68.) The rule given in Art. 66 conducts, in some casés, to 
negative exponents. 
Thus, let it be required to divide a° by a*.. We are directed 
to subtract the exponent of the divisor from the exponent of 
the dividend. We thus obtain 


a’—*=a-~, 


DHIOnrrh WW = 


3 
But a° divided by a® may be written de and since the value 
y y a 


C 


34 DIVISION. 


of a fraction is not altered by dividing both numerator and de- 
nominator by the same quantity, this expression is equivalent 


to —. 


2 


; ] 
Hence a—? is the same as = 


and these expressions may be used indifferently for each other. 
So, also, if a? is to be divided by a’, this may be written 


ape | 
That is, the reciprocal of a quantity is equal to the same quan- 
tity with the sign of its exponent changed. ' 


So, also, . Gout fn An ane ee 
De le © 
ad“ a 
And a hacia os 


(69.) Hence any factor may be transferred from the numer- 
ator to the denominator of a fraction, or from the denominator 
to the numerator, by changing the sign of its exponent. 


Thus, —=ab—. 





That is, the denominator of a fraction may be entirely re- 
moved, and an integral form be given to any fractional ex- 
pression. 

This use of negative exponents must be understood simply 
as a convenient notation, and not as a method of actually de- 
stroying the denominator of a fraction. Still this new nota- 
tion has many advantages, and is often employed, as will be 
seen hereafter. 

When the division can not be exactly performed, it may be 
expressed in the form of a fraction, and this fraction may be 


f DIVISION. 35 


reduced to its lowest terms, according to a method to be ex- 
plained in Art. 83. ° 

(70.) It frequently happens that the exponents of certain let- 
ters in the dividend are the same as in the divisor. 

Let it be required to divide a’ by.a’. The quotient is ob- 
viously 1, for every number is contained in itself once. But 
if we apply the rule of exponents, Art. 66, we shall have 


2—2 


we Por: 
Hence . a=): 

Again; let it be required to divide a” by 2%. Tne quotient 
is obviously 1, as before; and applying the rule of exponents, 
we obtain 

| ann" Ora" 
That is, every quantity affected with the exponent zero, is equal 
to unity. } 

This notation has the advantage of preserving a trace of a 
_ letter which has disappeared in the operation of division. 

Thus, let it be required to divide a*b* by a*b*. The quotient 
will be ab’. This expression is of the same value as a alone, 
and is commonly so written. If, however, it was important to 
indicate that the letter b originally entered into the expression, 
this might be done without at all affecting the value of the re- 
sult by writing it 

ab’. 

(71.) The proper sign to be prefixed to a quotient is readily 
deduced from the principles already established for multipli- 
cation. The product of the divisor and quotient must be equal 
to the dividend. Hence, — 


because +axX+b=+ab) . : +ab-+b=-+a. 
—aX +b=—ab } ) —ab++b=—a. 
herefe 
Besa a at BOD). oh ba, 
mate is pall -+-ab+-—b==a, 


Hence we have the following 


RULE FOR THE SIGNS. 
When both the dividend and divisor have the same sign, the 


quotient will have the sign +; when they have different signs, 
the quotient will have the sign —. 


36 DIVISION. 


EXAMPLES. 
Ex. 1. Divide —15ay’ by 3ay. “wy 
Ex. 2. Divide —18az’y by —9axr. +2 vy 
x. 3. Divide 150a*%be by —5ac. a5 9 ath 
Ez. 4. Divide 40a*b‘c by —abe. LY ah.” 

CASE II. 
(72.) When the divisor is a@ monomial, and the dividend a 
polynomial. - 


We have seen, Art. 51, that when a single term is multi- 
plied into a polynomial, the former enters into every term of 
the latter. 

Thus, a(a+b)=a@’+ab. 
Hence (a’*+ab)+a=a+b. 
Whence we deduce the following 


RULE. 
Divide each term of the dividend by the divisor, as in the for- 
mer case. 
EXAMPLES. 


Ex. 1. Divide 3x°+6x*+3ar—152 by 3z. 
Ans. a 


Ex. 2. Divide 3abc+12abx—9a’b by 8ab. =e + 4% | 
Ex. 3, Divide 4008 +600')"—17ab by ab, 9" foes +/7 “4 
Ex. 4. Divide 15a*bc—10acxu?+5ac*d* by —5a’*c. —3/ piv’ ye acl 
Ex. 5. Divide 6a*z*y*— 12a*x*y’+15a‘z*y* by 3a’x* uF a2} td ce Maia 
Ez..6. Divide 2° —2"°¥-+-2"h—27"™ wy, a tat x " 
Ex. 7. Divide 12a'y*’— 16a°y'+20a° y Meee hell — 4a ay 

CASE IIL ; wii 


(73.) When the divisor and dividend are both Fe si 

Let it be required to divide 2ab-+-a’+b* by a+b. 

The object of this operation is to find a third polynomial 
which, multiplied by the second, will reproduce the first. 

It is evident that the dividend is composed of alk the partial 
products arising from the multiplication of each term of the 
divisor by each term of the quotient, these products’ being add- 
ed together and reduced. Hence, if we can discover a term 


DIVISION. 37 


of the dividend which is derived without reduction from the 
multiplication of a term of the divisor by a term of the quo- 
tient, then ‘dividing this term by the corresponding term of the 
divisor, we shall be sure to obtain a term of the quotient. 

But from Art. 58, it appears that the term a’, which contains 
the highest exponent of the letter a, is derived, without reduc- 
tton, from the multiplication of the two terms of the divisor 
and quotient which are affected with the highest exponent of 
the same letter. Dividing then the term a’ by the term @ of 
the divisor, we obtain a, which we are certain must be one 
term of the quotient sought. Multiplying each term of the di- 
visor by a, and subtracting this product from the proposed 
dividend, the remainder may be regarded as the product of 
the divisor by the remaining terms of the quotient. We shall 
then obtain another term of the quotient by dividing that term 
of the remainder affected with the highest exponent of a, by 
the term a of the divisor, and so on. 

Thus we perceive that at each step we are obliged to search 
for that term of the dividend which is affected with the high- 
est exponent of one of the letters, and divide it by that term 
of the divisor which is affected with the highest exponent of 
the same letter. We may avoid the necessity of searching for 
this term by arranging the terms of the divisor and dividend 
in the order of the powers of one of the letters. 

The operation will then proceed as follows: 


The arranged dividend =a’+2ab+-b’|a+b= the divisor. 
a’+ ab a+b= the quotient. 
ab+b’= first remainder. 
ab+b? 
0 





It is generally convenient in Algebra to place the divisor on 
the right of the dividend, and the quotient directly under the 
divisor. 

(74.) From this investigation we deduce the following 


RULE FOR THE DIVISION OF POLYNOMIALS. 


1. Arrange the dividend and divisor according to the powers 
of the same letter. 


38 | DIVISION. 


2. Divide the first term of the dividend by the first term of the 
divisor, the result will be the first term of the quotient. 

3. Multiply the divisor by this term, and subtract the product 
from the dividend. 

4, Divide the first term of the remainder by the first term of 
the divisor, the result will be the second term of the quotient. 

5. Multiply the divisor by this term, and subtract the product 
from the last remainder. Continue the same operation till all 
the terms of the dividend are exhausted. 

If the divisor is not exactly contained in the dividend, the » 
quantity which remains .after the division is finished must be ~ 
placed over the divisor in the form of a fraction, and annexed | 


to the quotient. : 
EXAMPLES. 
1. Divide 2a*b+b’+2ab’?+a* by a’+b’+ab. 
. . Ans. a+b. 


2. Divide z*—a°+8a*x—8ax* by x—a. 
Ans. x’—2azr+a’. 
3. Divide a°+2°+2a°2* by a’—az+2’. 
Ans. a‘-+-a*z+az°+2", 
4, Divide a°—16a‘x’+642° by a’—4ax+42’. 
5. Divide a‘+6a’x’*—4a°x+2‘—4az* by a’—2ax+2". 
Ans. a’—2ar+2’. 
6. Divide z*+a°y’+y* by 2’+ayt+y. «© hs 2X) AGt “4 
7. Divide 127*—192 by 37-6. (ns 
Ans. eahibditig ie oak 
8. Divide 6z2°—6y° by 22°—2y*, Fad. Sat — Sata tS yf 
9. Divide a°+3a’b‘'—3a‘b’—b* by a *_ga°b-+-8ab" —b’. 
Ans. a°+38a*b+3ab’+0°. 
10. Divide a°—b* by a—b. t,4. Q+art 

11. Divide a*—b* by a—b. ; a°+ G20 F 

If the first term of the arranged dividend j is not divisible by 
the first term of the arranged divisor, the complete division is 
impossible. 

(75:) Hitherto we have Ane the terms of the quotient 
to be obtained by dividing that term of the dividend affected 
with the highest exponent of a certain letter. But, from the 
second remark of Art. 58, it appears that the term of the divi- 
dend affected with the lowest exponent of any letter is derived, 


/3 
& 


DIVISION. zi 39 
without reduction, from the multiplication of a term of the di- 
visor by a term of the quotient. Hence we may obtain a term 
of the quotient by dividing the term of the dividend affected 
with the lowest exponent of any letter, by the term of the di- 
visor containing the lowest power of the same letter, and 
nothing prevents our operating upon the highest and lowest 
exponents of a certain letter alternately in the same example. 

(76.) From the examples of Art. 74, we perceive that a°—b° 
is divisible by a-—-b; and a‘—O‘ is divisible by a—b. We shall 
find the same to hold true, whatever may be the value of the 
exponents of the two letters. That is, the difference of any 
two powers of the same degree is divisible by the difference of 
their roots. 


Thus, let us divide a°—b* by a—b. 





a’—b° |a—b 
a’—a'b\a’ 
a‘b—b*. 


The first term of the quotient is a*, and the first remainder 

is a*b—b*, which may be written 
b(a*—b*). 

Now if, after a division has been partially performed, the re- 
mainder is divisible. by the divisor, it is obvious that the divi- 
dend is completely divisible by the divisor. But we have al- 
ready found that a‘'—D* is divisible by a—b; therefore a°—b° 
is also divisible by a—b; and in the same manner it: may be 
proved that a°—D’ is divisible by a—b, and so on. 

To exhibit this reasoning in a more general form, let us 
represent any exponent whatever by the letter n, and let us 
divide a*—b" by a—b. 

a"—b” |a—b 
a*—ba"—|a"—" 

First remainder = ba’—'—b". 


_ Dividing a” by a, we have, by the rule of exponents, a*— for 
the quotient. Multiplying a—b by this quantity, and subtract- 
ing the product from the dividend, we have for the first re- 
mainder ba"—'—b", which may be written 

b(a**—b"). 
Now if this remainder is divisible by a—8, it is obvious that 
the dividend is divisible by a—b. That is to say, if the differ- 


40 € DIVISION. 


ence of the same powers of two quantities is divisible by their 
difference, the difference of the powers of the nzext higher de- 
gree is also divisible by that difference. ‘Therefore, since a*—b* 
is divisible by a—b, a’—b*° must be divisible by a—d; also, 
a°—b*, and so on. 

The quotients obtained by dividing the difference of the 
powers of two quantities by the difference of those quantities, 
follow a simple law. Thus, 

(a’—b*)+(a—b)=a+b. 
(a’—b*) +(a—b) =a +ab+0’. 
(a‘—b') + (a—b) =a +a°b+al?+0%. 
(a°—b’)+(a—b) =a +a +a°b' +ab’ +0. 
&c., &c., &e. 
(a"—b")~ (a—b) =a" +a bt eV +... AVF +0b 4. 

The exponents of a decrease by unity, while those of 0} in- 
crease by unity. 

(77.) It may also be proved that the difference of two even 
powers of the same degree is divisible by the sum of their roots. 

Thus, 

(a’—b*) + (a+b)=a—b. 

(a'—b') + (at+b)=a—vb+ab’—Dd’. 

(a°—b*) (a+b) =a —a’b+a'b’—a’b’+ab*—b', 
&ec., &c., &c. 

Also, the sum of two odd powers of the same degree is divisi- 
ble by the sum of their roots 

Thus, 

(a’ -+-b*) + (a+b) =a —ab+b’. 

(a? +0°) + (a+b) =a'—a*b+a°b?—ab' +0%. 

(a’ +b’) + (a+b)=a'—a'b+a'b?—a'b' +a°b'—ab’ +0. 
&c., &c., We. 

(78.) The preceding principles will enable us to resolve va- 
rious algebraic expressions into their factors. 

1. Resolve a*—D’ into its factors. . 

Ans. (a'--ab+0) (a=), 


2. Resolye a*+0’ into its factors. rdf AOL + OR) (a +o) i 
3. Resolve a°—D° into four factors. ea Labios hx f hay 4 td 
4, Resolve a°—8b° into its factors, rs (@+<wT 7 pA ©) 
5. Resolve 8a°—1 into its factors. - ., [4 Pr kury) (La—/ , 
6. Resolve 8a*—8b" into three factors, “7444 bpd b ) (2a -26) 


DIVISION. 4] 


7. Resolve 1+27b° into its factors. 7rd (VAS YY ° H 

8. Resolve 8a°+27" into its factors, . Hw. (Ae €el/ 70 / 

(79.) One polynomial can not be divided by “saglive poly- 
nomial containing a letter which is not found in the dividend ; 
for it is impossible that one quantity multiplied by another 
which contains a certain letter, should give a product not con- 
taining that letter. 

A monomial is never divisible by a polynomial, because 
every polynomial multiplied by another quantity gives a prod- 
uct containing at least two terms not susceptible of reduction. 

Yet a binomial may be divided by a polynomial containing 
any number of terms. . 

Thus, a*—b* is divisible by a°+a°b+ab’+b*, and gives for a 
quotient a—b. 

_ So, also, a binomial may be divided by a polynomial of a 
hundred terms, a thousand terms, or, indeed, any finite num- 
ber. 


DIVISION BY DETACHED COEFFICIENTS. 


(80.) We have shown, in Art. 64, how multiplication may 
sometimes be conveniently performed by operating upon the 
coefficients alone. The same principle is applicable to divi- 
sion. ‘Thus, take the example of Art. 73, to divide a’+2ab+b’ 
by a+b; we may proceed as follows: 

£4-2+1)1+1 

1+1  |1+1 
1+1 
1+1 

The coefficients of the quotient are 1+1. Moreover, a’~a 
=a; and therefore ‘a is the first term of the quotient, and b the 
second. 

Ez. 2. Divide x*—3az’—8a’x’ + 18a*°x—8a' by x? +2axr—2a’. 

1—8— 8+18—8/1+2—2 


14+2— 2 1—5+4 
—5— 6+18—8 . 
= § 10-410 

4+ 8—8 

4+ 8-8 


The coefficients of the quotient are 1—5+4, and it remains to 


42 DIVISION. 44 


supply the letters. Now z*+2’?=z’; and a‘+a’=a’*. Hence 
x’, ax, and a’ are the literal parts of the terms, and’ therefore 
the quotient is 
x’ —5ax+4a’. 

Ez. 3. Divide 6a*—96 by 3a—6. 

Here, as we have the fourth power of a without the lower 
powers, we must supply the coefficients of the absent terms, 
as in multiplication, with zero. 


6+ 0+0+0—96/3—6 





6—12 2+4+8+16 
12 
12—24 
24 
24—48 
48— 96 
48—96 
But a‘~a=a’; hence the quotient is 
2a*+4a*+8a+16. 
Ex. 4. Divide 3y°+3zry?—4z°y—42° by z+y. 
Ans. 3y°—42". 


Ex. 5. Divide 8a°—4a*z—2a°x*+a°x* by 4a’—2’. 
Ans. 2a°—a’z. 
Ez. 6. Divide a°+4a°—8a'—25a'+35a?+2la—28 by a+ 
5a+4. 
Ans. a'—a’—a’4+14a—7. 


SECTION VI. 


FRACTIONS. 


(81.) When a quotient is expressed as described in Art. 16, 
by placing the divisor under the dividend with a line between 
them, it is called a fraction ; the dividend is called the numer- 
ator, and the divisor the denominator of the fraction. Alge- 
braic fractions do not differ essentially from arithmetical frac- 
tions, and the same principles are applicable to both. 

The following principles form the basis of most of the oper- 
ations upon fractions: 

1. In order to multiply a fraction by any number, we must 
multiply its numerator, or divide its denominator by that num- 
ber. : 


Thus, the value of the fraction s is &. If we multiply the 


ee) , ay 
numerator by a, we obtain ra ab; and if we divide the de- 


nominator of the same fraction by a, we obtain also ab; that 

is, the original value of the fraction b has been multiplied by a. 
2. In order to divide a fraction by any number, we must di- 

vide its numerator or multiply its denuminator-by that number. 


« 2b 
Thus, the value of the fraction — is ab. If we divide the 
numerator by a, we obtain a or b; and if we multiply the de- 


» ab 
nominator of the same fraction by a, we obtain ss or b; that 


is, the original value of the fraction ab has been divided by a. 
3. The value of a fraction is not changed if we multiply or 
divide both numerator and denominator by the same number. 


A" 


44 FRACTIONS. 


ab_abx aby 


Thus, 





a ax axy 

Every quantity which is not expressed under a fractional 
form, is called an entire quantity. 

An algebraic expression composed partly of an entire quan- 
tity and partly of a fraction, is called a mixed quantity. 

(82.) The proper sign to be prefixed to a fraction may be 
determined by the rules already established for division. The 
sign prefixed to the numerator of a fraction affects merely the 
dividend; the sign prefixed to the denominator affects merely 
the divisor; but the sign prefixed to the dividing line of a 
fraction a bed the quotient. 


Thus, — C= +b, for + divided by + gives +. 


so) 


—b, for — divided by + gives — 


ab 


ab__», for + divided by — gives — 


Sa 48, for — divided by — gives +. 


ab ’ | 
So, also, ne ger CAs for this shows that the former quotient 


b is to be subtracted, which is done by changing its sign. 


—ab 
— bie because the former quotient —b is to be 


subtracted, whence it becomes +8. 


ab 
ABIL for the same reason ; 





and Lolli My also for the same reason. 
Hence we have the following equivalent forms: 
ab —ab —ab ab 
<r — +5 : 
a —a a —a 
—ab ab ab —ab 
also, ——_=——_ =——__ =~) 
aq —a a —a 


That is, of the three signs belonging to the numerator, de- 
nominator, and dividing line of a fraction, any two may be 
changed from + to — or from — to +, without affecting the 
value of the fraction. 


FRACTIONS, 45 


In the examples of fractions here employed for illustration, 
both numerator and denominator have consisted of monomials. 
The same principles are applicable to polynomials; but it 
must be remarked, that by the sign of the numerator we un- 
derstand the entire numerator as distinguished from the sign 
of any one of its terms taken singly. 

at+b-+c. —a—b—c 

Thus, oe ae equal to ee TTS 

When no sign is prefixed either to the terms of a fraction or 
to its dividing line, + is always to be understood. 


REDUCTION OF FRACTIONS. 
PROBLEM I. ? 


(83.) To reduce a fraction to lower terms. 


RULE. 
Divide both numerator and denominator by any quantity 
which will divide them both without a remainder. 
According to Remark 3, Aré. 81, this will not change the 
value of the fraction. 


ax a 
Thus, Pgs 3 
Also saat (dividing both numerator and de- 
f 5a’b* 5b = 
nominator by a’b.) 
ax" ax 


And PE ha Pie 

If the numerator and denominator are both divided by their 
greatest common divisor, it is evident the fraction: will be re- 
duced to its lowest terms. The method of finding the greatest 
common divisor is considered in Section X V.; but im the fol- 
lowing examples the greatest common divisor is easily found, 
by resolving the quantities into factors according to methods 
already indicated. 


EXAMPLES. 
: cx +2? 

1, Reduce ———— to its. lowest terms. 
actaz 


H 93 
Ansi — 
an 


46 FRAOTIONS. 


eal b 








2. Reduce ; a haneehe to its lowest terms. 
oe 
Ans. Be’ 
z*—a’ . 
3. Reduce — to its lowest terms. 
x*—a'* 
] 
Ans zh 
22°— 162—6 
; _—______ to its lowest terms. 
4, Reduce 37) Leap 10S 
7 Ans pas? 
3x°y+3zy’ 
. Reduce ——-_ to its lowest terms. 
5. Reduce Ba" + Oxy -+3y i 
* xy 
Ans. ——. 
LY 
a—b eee 
6. Reduce 7 F_oab LP to its lowest terms. bid TD 
(Me |, CPARTE 
7. Reduce -————— to its lowest terms. 4 ae 
Mi LOLA . 


PROBLEM II. 
(84.) To reduce a fraction to an entire or mixed quantity. 


RULE. 


Divide the numerator by the denominator for the entire part, 
and place the remainder, if any, over the denominator for the ” 
fractional part. 


27 
Thus, eae 5= 52, 


Also, E 





+2" a 
=(az+a’)+z=a+-.: 

x 

EXAMPLES. 


ax—x* : 
1. Reduce to an entire quantity. 





FRACTIONS. 47 








b— 2a? : 
2. Reduce ——— to a mixed quantity. 
a 2 ‘ ; 
3. Reduce = to a mixed quantity. 
22° 
Ans, a-+-a+ ; 
a—x 


z 
4, Reduce amy to an entire quantity. 


10z*—5z+3 ; 

5. Reduce Ss to a mixed quantity. 
8b°— 16b+7a’b? : 

6. Reduce ee to a mixed quantity. 


PROBLEM III. 
‘ — (85.) To reduce a mixed quantity to the form of a fraction. 
RULE. 


Multiply the entire part by the denominator of the fraction ; 
to the product add the numerator with its proper sign, and place 
the result over the denominator. 


8X54+2 15+2 17 














2— = 
Thus, 32 ; ; 5 
This result may be proved by the preceding Rule. For 
Pree 
b aXc+b act+b 
Also, Reet eal 2 : 
C C c 
EXAMPLES. 
a—x’ i 
1. Reduce z+ to the form of a fraction. 
Ans oes 
x 
2 
2. Reduce 2a 3 to the form of a fraction. 
8az+2’ 








48 FRACTIONS, 

2 —7 ee 7 al 
3. Reduce ot the form of a fraction. «4. 5 

x—a—l " Jd. at, 
4, Reduce 1+ — to the form of a fraction» oo A Ca 

<a z—3 . AS 6%-+4 | ox?- a 
5. Reduce 1+ee+— to the form of a fraction. . res 

~ 8c? D4 Far 32 £28 1% 

6. Reduce Ebest to the form of a fraction: rae, 


—p ee 


PROBLEM IV. 


(86.) To reduce fractions to a common denominator. 


RULE. 


Multiply each numerator into all the denominators, except its 
own, for a new numerator, and all the denominators together for - 
a common denominator. 


EXAMPLES. 
be: Cnxcngd . ; 
I. Reduce b and 7 to a common denominator. 
ad be 
"bd? ba’ 
Here it ‘will. be seen that the numerator and denominator of 
the first fraction are both multiplied by d, and in the second 
fraction they are both multiplied by b. The value of the frac- 
tions, therefore, is not changed by this operation. 


Ans 


a a+b ' : ; 
2. Reduce b and — to equivalent fractions having a com- 


mon denominator. 


ac ab-+b’ 
Anse 3-3; —g—. 
eS Be be 
3x 2b ; 
3. Reduce aa) et and d to fractions having a common de- 
nominator. | f a : roe 


S 2 : 
4, Reduce -, —, and laty to fractions having a common 


s 4 3 al weg IB | GE 
denominator. Se) Pon, 


FRACTIONS. 49 


32 
5. Reduce — rs yy and <= = to. inp. opp nes having a a common 
denominator. Me San Se , 
zx x+l 1—2z 
6. Red —, ——, 
e uce 9° 5 and —— Re Fe to fractions having. a common 
denominator. La, ‘ 


4 bs . ° 
to fractions having a common de- 





7. Reduce — = ; and on 
3a? 4x 
nominator. 

Following the Rule, we obtain 

Sax 3ax’+6z* 

— and aE Pa 

122° Lore : 
which fractions have a common denominator, and are equiva- 
lent to those originally proposed. Nevertheless, it may be ob- 
served, that these fractions are not reduced to their /east com- 
mon denominator, for every term is divisible by z. The least 
common denominator is the least common multiple of the de- 
nominators of the proposed fractions. 

A common multiple of two or more numbers is any number 
which they will divide without a remainder ; and the /east com- 
mon multiple is the least number which they will so divide. 
Thus, 122° is the least common multiple of 32? and 42; and 
the above fractions reduced to their least common denomina- 
tor are 

8a 3ax+6x° 

and. 

13z5 122° 
The least common multiple of two numbers is their product 
divided by their greatest common divisor. 


3 5 ; : : 
8. Reduce i and a '° equivalent fractions having the least 


common denominator. 
The product of the denominators is 294, which, divided by 
7 (their greatest common divisor), gives 42, the least common 
denominator, and the required fractions are 
10 


9 
qa 2nd 35° 


7 11 
9, Reduce the fractions in and is to others cans have the 


least common denominator. 


D 


50 FRACTIONS. 


bd 
10. Reducei=— ah and =; ape t° equivalent fractions having the 


least common denominator. 


4ac d 
Ans. 6bc? and rc 
b d ; : : 
11. Reduce — and cates to equivalent fractions having 
the least common ae ORES 








PROBLEM V.. 
(87.) To add fractional quantities together. 


RULE. 


Reduce the fractions to a common denominator ; add the nu- 
merators together, and place their sum over the common denom- 
inator. 

The fractions must first be reduced to a common denomina- 
tor to render them like parts of unity. Before this reduction, 
they must be considered as unlike quantities. 

e 
EXAMPLES. 


x x 
1. What is the sum of ¥ and ay 


Reducing to a common denominator, the fractions become 
3x 22 


G and |. 


. pe 
Adding the numerators, we obtain Se" 


‘It is plain that three sixths of x and two sixths of 2 make 
five sixths of z. 


é 
2. Required the sum of 4 and 7 








adf +-bef+-bde 


Ans baf 


FRACTIONS. 51 


qo 


a+b a—b 


2a p deci Tae. 
gr One a 


a 


. Required the sum of 52, 


\ 


; 2 8 
. Required the sum of 2a, sat—, and at—. \ 


or 


| 58x 
Ans. bet —e- 


a—ZX 


e 





: a 
Required the sum of a+za, Se and 


Var se woudl 








a’ —ax 
7. Required the sum of nt and oe. 
Ans. a. 
. a a—2m at+2m 
8. Required the sum of Bit i and row 
—b b / 
9, Required the sum of end oi . 





an 4 
mtn m+n 


PROBLEM VI. 
(88.) Lo subtract one fractional quantity from another. 


RULE. 


Reduce the fractions to a common denominator, subtract one 
numerator from the other, and place their difference over the 


common denominator. 


EXAMPLES. 


22 3x 
1. From Pe subtract —. 


5) 
Reducing to a common denominator, the fractions become 
10z 9z 
Teer ats 
LOn OO T 
Hence is 15 15 3 


and it is plain that ten fifteenths of z, diminished by nine fit- 
teenths of 2, equals one fifteenth of zx. 


: a b 4 
. Required the sum of —— and rd, 


52 FRACTIONS. 


; 1 pe ar 
2. From — subtract oh 





Bo] anc 3 Fae 
aot subtract — wad : fi) BE “ aa 
It must be remembered, that the minus sign before the di- 
viding line of a fraction affects the quotient (Art. 82); and 
since a quantity is subtracted by changing its sign, the result 
of the subtraction in this case is 


9x 
3. From 





9x—4y 5x—3y 

ea P 
which fractions may be reduced to a common denominator, 
and the like terms united, as in addition. 











ax 
4, From ——— — — subtract —— Fie 
2 
Ans. eri 
b’—c 
24-7x 5z2—6 
5. From arp subtract z— 2 a 
355z2—6 
Ans 168 
a —a 7 
6. From Ba+5, subtract be sh 
b a—b 
7. From and subtract ——. 
4 J 
13a— 5b —2Qb 
8. From ats subtract — 
25a—11b 
A e a ae 
deen” (5 


PROBLEM VI. 
(89.) To multiply fractional quantities together. . 


RULE. 


Multiply all the numerators together for a new numerator, - 
and all the denominators together for a new denominator. 


Let it be required to multiply 5 by > 


FRACTIONS. 53 
First, let us multiply ; by c. According to Remark first of 
ac 
Art. 81, the product must be S° 


But the proposed multiplier was =; that is, we have used a 


* 
multiplier d times too great. We must therefore divide the 


result + by d; and, according to Remark second of Art. 81, 


we obtain 
ac 
bd’ 


which result conforms to the Rule above given. 





EXAMPLES. 
v 2x 
1. Multiply 6 by Ts 
a 
Ans. oT 
10z 
2. What is the continued product of 5 —, and oT 
3. aa 
4. What is the continued product of e— og b and a 
Ke 9ax. 
b 
5. Multiply re by <. 
z+ 6" 
6. Multiply = _—— youl Pg. | 
x‘*—b* 
a b’c-+be 
7. What is the continued product of 2, —— tt , and tig bg 
a a+b - 
Ans er” 


54. FRACTIONS. 


ae a—b 
Sp a+b 





8 Multiply <, 
a’ +b? 


An Scie epiiya 


: 1 ] 
(90.) Ez. 1. Multiply - by = 
‘ l 
According to the preceding Article, the result must be 7 
1 1 
But, according to Art. 68, 7 may be written a; a may be 


; 1 
written a—’; and 7s may be written a~. 


Therefore, a? xXa"=a—. 

That is, the Rule of Art. 50 is general, and applies to nega- 
tive as well as positive exponents. 

_ Ex. 2. Multiply —b~ by b~. 
Ans. —b-. 

. Multiply a— by a’. 
. Multiply b— by 0°. 
. Multiply a by a—”. 
. Multiply 6 by 0”. . 
. Multiply (a—b)* by (a—b)—. 


IOP Ww 


PROBLEM VIL. 
(91.) To divide one fractional quantity by another. 


RULE. 


Invert the divisor, and proceed as in multiplication. 

If the two fractions have the same denominator, then the 
quotient of the fractions will be the same as the quotient of 
their numerators. 


A ; 3. ag 
Thus it is plain that i238 contained in Ts 28 often as 3 is 


contained in 9. 
But when the two fractions have not the same denominator, 
we must reduce them to this form by Problem IV. 
a 


Let it be required to divide j by = 


=a 


FRACTIONS. 2 5a 


Reducing to a common denominator, we have S to be di- 
be 
vided by bd 
It is now plain that the quotient must be represented by the 
division of ad by bc, which gives 
ad 
be 
the same result as obtained by the above Rule. 
ad lag d_ad 








2ani8; PCr EOE a 
EXAMPLES. 
1. Divide = = by = 
Ans. 14. 
Loe. 2a 4c Ag 
2. Divide 7% by et es a ee 
3. Divid Pian b 7 
. Divide Pix anon 
1 
MERvide' = by = 
: b Ms Yor 
5. Divide Bcd by rr 
x—b 
An 6c’x 
6. Divide ante by = 
—x c—2zx 
2a+z 
Ans Cc +cxr +2 
b a b 
7. Divide ate a pa hale 


Ans. Unity. 
Lee | 1 
(92.) Hz. 1. Divide z by = 


According to the Rule of the preceding Article, we have 
a ae | 


56 FRACTIONS. 


But ie may be written a; es may be written a; and 5 
is equal to a~ 

Hence a =a =a + 

That is, the Rule of Art. 66 is gcc and applies to nega- 
tive as well as positive exponents. 

Ez. 2. Divide —b~ by —b—. 

Ans. b-, 

. Divide a’? by a“ 
. Divide 1 by a~. ni, 
. Divide 6a" by —2a~ Fut, —/ha™ 
. Divide b"— by 6”. ; iG 
. Divide 122—~y—* by —4zy*. -” —HEx 

8. Divide (x—y)— by (x@—y)—. L We 

(93.) According to the definition, Art. 33, He ee, of 
a quantity is the quotient arising from dividing a unit by that 
quantity. 


IO Oo PP CO 


Hence the reciprocal of ; 


b b 
is 1+ =1 X-=-: 
b aa 
That is, the reciprocal of a fraction is the fraction inverted. - 
DAE 


F Axe, 
Thus the reciprocal of a ein 


ins 
The reciprocal of is b+e. 
Hence, to divide by any quantity is the same as to multiply 
by its reciprocal, and to multiply by any quantity is the same 


as to divide by its reciprocal. 
(94.) The numerator or denominator of a fraction may be 


itself a fraction ; 
a 
As b or 
c 


Such expressions are easily reduced by applying the pre- 
ceding principles. 


; Se 
Be 


7 ag. 


FRACTIONS. 
ae: i a 
Thus, io) means ae 
c 
which, according to Remark second, Art. 81, equals 7 
% 
} aa We 
Again, (*) means'a——; 


which, according to Art. 91, equals - 
a 
iso. — ( means the same as @.. © 
he 
) 


which, according to Art. 91, equals ‘e 


Ez. 1. Find the value of the fraction 


Ex. 2. Find the value of the fraction 


QS ail [esjto 
09 ic 4 


SECTION VII. 


SIMPLE EQUATIONS. 


(95.) An equation is a proposition which declares the equality 
of two quantities expressed algebraically. 

Thus, z—4=b—z, is a proposition expressing the equality 
of the quantities e—4 and b—z. 

The quantity on the left side of the sign of equality is called 


the first member of the equation; the quantity on the right, the 


second member. 

Equations are usually composed of certain quantities which 
are known, and others which are unknown. The known quan- 
tities are represented either by numbers or by the first letters 
of the alphabet, a, b,c, &c.; the unknown quantities by the last 
letters, x, y, z, &c. 

An identical equation is one in which the two members are 
identical, or may be reduced to identity by performing the op- 
erations which are indicated in them. 

Thus, 2Qa—5=2r—5 

32 +42=7Tx 
(ety) (@—-y)=2*—-y. 

A root of an equation is the value of the unknown quantity 
in the equation. 

(96.) Equations are divided into degrees, according to the 
highest power of the unknown quantity which they contain. 

Those which contain only the first power of the unknown 
quantity are called simple equations, or equations of the first 
degree. 

As ax+b=cxr-+d. 

Those in which the highest power of the unknown quantity 


Hua 


SIMPLE EQUATIONS. 59 


is a square, are called quadratic equations, or equations of the 
second degree. 

As 473 —=22=5—2". 

Those in which the highest power is a cube, are called cubic 
equations, or equations of the third degree. 


As 2+ px =2¢: 

So, also, we have biquadratic equations, or equations of the 
fourth degree ; equations of the fifth, sixth, —— — — — nth 
degree. 

Thus, z"+-p2"—'=r, is an equation of the nth degree. 


In general, the degree of an equation is determined by the 
highest of the exponents with which the unknown quantity is af- 
fected. 

* (97.) Numerical equations are those which contain only par- 
ticular numbers, with the exception of the unknown quantity, 
which is always denoted by a letter. 

Thus, 2°+4a°=32+12 is a numerical equation. 

Interal equations are those in which the known quantities 
are represented by letters, or by letters and numbers. 

Thus, z2°+-p2"+qz=r 

x*—3pa'+ 5gu'=5 

To solve an equation is to find the value of the unknown 
quantity, or to find a number which, substituted for the un- 
known quantity in the equation, renders the first member 
identical with the second. ; 

The difficulty of solving equations depends upon their de- 
gree, and the number of unknown quantities. We will begin 
with the most simple case. 


are literal equations. 


SIMPLE EQUATIONS CONTAINING BUT ONE UNKNOWN QUAN- 
BY, 

(98.) The various operations which we perform upon equa- 
tions in order to deduce the value of the unknown quantities, 
are founded upon the following principles: 

1. If to two equal quantities the same Suey be added, the 
sums will be equal. 

2. If from two equal quantities the same quantity be sub- 
tracted, the remainders will be equal. 

3. If two equal quantities be multiplied by the same quan- 
tity, the products will be equal. 


60 SIMPLE EQUATIONS. 


4, If two equal quantities be divided by the same quantity, 
the quotients will be equal. 

(99.) The unknown quantity may be combined with the 
known quantities in the given equation by the operations of 
addition, subtraction, multiplication, or division. : 

We shall consider these different cases in succession. 

I. The unknown quantity may be combined with known 
quantities by addition. 

Let it be required to solve the equation 

| . r+6=24., 

If from the two equal quantities, +6 and 24, we subtract 
the same quantity 6, the remainders will be equal, according 
to the last Article, and we shall have 7 

r+6—6=24—6, be 
or 7=24—6, 
=18, the, value of x required. 
So, also, in the equation 

z+a=h, 
subtracting a from each of the equal quantities, +a and }, the 
result is 

x=b—a, the value of x required. 

(100.) II. The unknown quantity may be combined with 
known quantities by subtraction. 

Let the equation be 

r—6=24. 

If to the two equal quantities, c—6 and 24, the same quan- 
tity 6 be added, the sums will be equal, according to Art. 98, 
and we have ) 

z—6+6=24+6, 
or x=80, the value of x required. 
So, also, in the equation 
x—a=b, 
adding a to each of these equal quantities, the result is 
x=b-+a, the value of z required. 

From the preceding examples,.it follows that 

We may transpose any term of an equation from one member 
to the other by changing its sign. 

We may change the sign of every term of an equation with- 
out destroying the equality. 


SIMPLE EQUATIONS. 61 


This is, in fact, the same thing as transposing every term in 
each member of the equation. | 

If the same quantity appear in each member of the equation 
affected with the same sign, it may be suppressed. 

(101.) III. The unknown quantity may be combined with 
known quantities by multiplication. 

Let the equation be 

62= 24. 

If we divide each of the equal quantities, 62 and 24, by the 

same quantity 6, the quotients will be equal, and we shall have 


=4, the value of x required. 
So, also, in the equation 
QL 
dividing each of thesé equals by a, the result is 


r=", the value of x required. 


From this it follows, that 

When the unknown quantity is multiplied by a known quan- 
tity, the equation is solved by dividing both members by this 
known quantity. 

(102.) IV. The unknown quantity may be combined with 
known quantities by division. 

Let the equation be 


If we multiply each of the equal quantities, = and 24, by the 


same quantity 6, the products will be equal, and we shall have 
x=144, the value of x required. 
So, also, in the equation 
ape 
multiplying each of these equals by a, the result is 
x=ab, the value of a required. 
From this it follows, that 
When the unknown quantity is divided by a known quantity, 


rN SR a ee ee 


\ 


62 SIMPLE EQUATIONS. 


the equation is solved by multiplying both members by this known 
quantity. 
(103.) V. Several terms of an equation may be Frackone 
Let the equation be 


zr 24 

QNOR His 
Multiplying each of these equals by 2, the result is 

Peds 38 

t= 31 e 

Multiplying each of these last equals by 3, we obtain 

omni 24 . | 
st 4-+—— 3 
and multiplying again by 5, we obtain 
1liz= 20 +24, 
an equation free from fractions. 

We might have obtained the same result by multiplying the 
original equation at once by the product of all the denom- 
inators. 

Thus, multiplying by 2X35, we have 

302 60 120 
eyes NG. 
or reducing, we have 
15z=20-+ 24, as before. 
So, also, in the equation 
Bro Mil 
ace’ 
multiplying successively by all the denominators, or by ace 
at once, we obtain 
acer abce  acde 





a c € 
Canceling from each term the letter which is common to its 
numerator and denominator, we have 
cex—=abe-+acd, 
an equation clear of fractions. - 
Hence it appears that 
An equation may be cleared of fractions by multiplying each 
member into all the denominators. 
(104.) From the preceding remarks, we deduce the fol- 
lowing 


\ 
\ 
- 


SIMPLE EQUATIONS. 63 


RULE FOR THE SOLUTION OF A SIMPLE EQUATION CONTAINING 
ONE UNKNOWN QUANTITY. 

1. Clear the equation of fractions, and perform in both mem- 
bers all the algebraic operations indicated. 

2. Transpose all the terms containing the unknown quantity 
to one side, and all the remaining terms to the other side of the 
equation, and reduce each member to its most simple form. 

3. Divide each member by the coefficient of the unknown 
quantity. 


EXAMPLES. 


1. Given 52+8=427-+10, to find the value of z. 
Transposing 42 to the first member of the equation, and 8 
to the second member, taking care to change their signs (Ar¢. 
100), we have 
52—4x=10—8. 
Uniting similar terms, x=2. 
In order to verify this result, put 2 in the place of 2 wher- 
ever it occurs in the original equation, and we shall obtain 
5X2+8=4X2+10. 
That is, 10+8=8+10, 
or IS=fs, 
an identical equation, which proves that we have found the 
correct value of z. ; 
2. Given r—1=2 +, to find the value of z. 
Multiplying every term of the equation by 5 and also by 8, 
in order to clear it of fractions (Art. 103), we obtain 
15a—105=382+ 52. 
Hence, by transposition, 


152z—32x—5zx=105, 


or Li 106; 
105 
and therefore r= == 15; 


To verify this result, put 15 in the place of in the original 
equation, and we have 


15 pen ate 
im thi Bo 


64 SIMPLE EQUATIONS. 


That is, 15—7=38+5, 
or 8=8, 
an identical equation. 
8. Given 3ar—4ab= 2ax—Gac, to find the value of z in ~ 
terms of b and c. 
Dividing every term by a, we have 
3x —4b=2x—6ce. 
By transposition, 
3x2 —2x4=4b—6e, 
or x=4b—6c. 
This result may be verified in the same manner as the pre- 
ceding. 
4. Given 8z7*—10x=8z+2’, to find the value of z. 











Ans. x=9. 
5. Given ue pid nenue pee to find x. 
dx d 
Ans. ve 
Cc 
— 8 
6. Given A se to find z. 
Ans. x=9. 
ab 1 
7. Given ed tas to find z. 
ab—1 
Ans. mame SWE 
6 llz—37 
8. Given 8z+ = =54— to find z. 
Ans: za-7%. 
9. Given 5ax—2b+4br=227-+5c, to find z. 
igo og (Sadek 
Me PS Baa ano 
37 — a4 
10. Given an Dena ca to find the value of z. 
Ans. x=5. 
3 a i, _ 97—7. 
11. Given 214+—— rites 0 —— ate BAN 5 us to find z. iy 


(105.) An ha may always be cleared of fractions by 
multiplying each member into a// the denominators according 


=» 


SIMPLE EQUATIONS. 65 


to Art. 103. But sometimes the same object may be attained 
by a less amount of multiplication. 

Thus, in the preceding example, the equation may be cleared 

of fractions by multiplying each term by 16, instead of 16X8 

2, and it is important to avoid all useless multiplication. In 

general, it is sufficient to multiply by the least common multiple 

of all the denominators. See Art. 86. 
12. Given 82-4 to find z. eh ai ag 
at+z .b—z 


18. Given 32— atco=————, to find z. 





4a°— 3b 
8a+38ac—3° 





Ans. x= 


14. Given ete aac ts to find z. 


b d’ 
abcd 


Ans. ¢= 357+ ad—4abd—2ab 


15. Given (a+z) (b+2)—a(bt0)=—" +2, to find z. 


ac 
Ans. x= 











5° 
17—3 ae 7. 4 
16. ade FANT H6e- eS, forfiid &. 
ey 3 
’ 3z2—3 20—2 627-8 47—4 
; 17. Given He RTE AR io aE UAT TO 5 to find z. 
v : 16 8 
18. Given me PE cheer . to find z. ceo 





21 Ac eclinig! 


. 62+7 Ta—138 _2et4 
19. Given ——— de + 


9 6z+3 “ag , to find x. 





5 yi 3 
20. Given ab-+zac—=cx=2a0-+2ab—Ger, to find the value 


of z. 
70ab—8ac 
saitah vist MEE 


SOLUTION OF PROBLEMS. 


(106.) The solution of a Problem by Algebra consists of 
two distinct parts : 


E 


66 SIMPLE EQUATIONS. 


1. To express the conditions of the problem algebraically ; 
that is, to form the equation. 

2. To solve the equation. 

The second operation has already been explained, but the 
first is often more embarrassing to beginners than the second. 
Sometimes the statement of a problem furnishes the equation 
directly ; and sometimes it is necessary to deduce from the 
statement new conditions, which are to be expressed alge- 
braically. The former are called explicit conditions ; and those 
which are deduced from them, implicit conditions. 

It is impossible to give a general rule which will enable us 
to translate every problem into algebraic language. The 
power of doing this with facility can only be acquired by re- 
flection and practice. F 

The following directions may be found of some service. 

Denote one of the required quantities by x; then, by means 
of this letter, with the algebraic signs, perform the same opera- 
tions which would be necessary to verify its value if it was al- 
ready known. 

Problem 1. What number is that, to the double of which if 
16 be added, the sum is equal to four times the required num- 
ber ? 

Let x represent the number required. 

The double of this will be 2z. 

This increased by 16 should equal 42. 

Hence, by the conditions, 27+16=4z2. 

The problem is now translated into algebraic language, and 
it only remains to solve the equation in the usual way. 

Transposing, we obtain 

16=427—227=22, 
and S27, 
or xz=8. 

To verify this number, we have but to double 8, and add 
16 to the result ; the sum is 32, which is equal to four times 8, 
according to the conditions of the problem. 

Prob. 2. What number is that, the double of which exceeds 
its half by 6? 3 

Let x = the number required. 

Then, by the conditions, 


es 


SIMPLE EQUATIONS. 67 


pgees 
2r—5=6. 
Clearing of fractions, 
4x—x=12, 
or dr= 12. 
Hence r=4, 


To verify this result, double 4, which makes 8, and diminish 
it by the half of 4, or 2; the result is 6, according to the con- 
ditions of the problem. 

Prob. 3. The sum of two numbers is 8, and their difference 
2. What are those numbers? 

Let z = the least number. 

Then «+2 will be the greater number. 

The sum of these is 27+2, which is required to equai 8. 
Hence we have 


27+2=8. 
By transposition, 2x—8—2=6, 
and x=8, the least number. 
Also, x+2=5, the greater number. 
erification. 5+3=8 - 
aut a 3—9 | according to the conditions. . 


The following is a generalization of the preceding Problem. 

Prob. 4. The sum of two numbers is a, and their difference 
b. What are those numbers? 

Let x represent the least number. 

Then «+6 will represent the greater number. 

The sum of these is 2x-+b, which is required to equal a. 

Hence we have 


22+b=a. 
By transposition, 272=a—b, 
a—b a b 
on a the less number. 


b b 
Hence r+b=5—5+b=5+5, the greater number. 


As these results are independent of any particular value at- 
tributed to the letters a and b, it follows that 

Half the difference of two quantities, added to half their sum, 
is equal to the greater; and 


68 SIMPLE EQUATIONS, 


Half the difference subtracted from half the sum is equal to 
the less. 


MAA A ae. a b 
The expressions alps and 379 are called formulas, because 


they may be regarded as comprehending the solution of all 

questions of the same kind; that is, of all problems in which 

we have given the sum and difference of two quantities. 
Thus, let a=8 


a in the preceding problem. 
b 8+2 . 
Then sto o, the greater number. 
1 b 8—2 
And eg pe the less number. 
~ = 10; their difference = 6; required the numbers.- ‘ 
5 8 12 ; 2 “ es 
Mi & 93 66 ll 6é ve 
2) 
i= 100 « 50 “ 73 
5 = 100 “ 1 “ ‘0fe% 
= S 5 13 i ‘“ DY 
oe i ' ‘ 
10 1 


Prob. 5. From two towns which are 54 miles distant, two 
travelers set out at the same time with an intention of meet- 
ing. One of them goes 4 miles and the other 5 miles per hour. 
In how many hours will they meet? 

Let x represent the required number of hours. | 

Then 42 will represent the number of miles one traveled, 

and 5x'the number the other traveled ; 
and since they meet, they must together have traveled the 
whole distance. 


Consequently, 4x+5x=54. 
Hence 92-54, ‘ 
or x=6. 


Proof. In 6 hours, at 4 miles an hour, one would travel 24 
miles; the other, at 5 miles an hour, would travel 80 miles. 
The sum of 24 and 30 is 54 miles, which is the whole distance. 

This Problem may be generalized as follows: 

Prob. 6. From two points which are a miles apart, two 
bodies move toward each other, the one at the rate of m miles 


SIMPLE EQUATIONS. 69 


per hour, the other at the rate of x miles per hour. In how 
many hours will they meet? 

Let-x represent the required number of hours. 

Then maz will represent the number of miles one body 
moves, 

and nz the miles the other body moves, 
and we shall obviously have 


mx+nx=a. 
a 
mt-n 





Hence x= 


This is a general formula, comprehending the solution of all 
problems of this kind. Thus, 


—150; ~ 6; y 4 S 

o o ? a 7 o 
BrerO Wise yn S gare OT BS 
Yes 135 Ae 15 12 are 
ag = o Dy 
210 5 20 S 15 2 


Required the time of meeting. 

We see that an infinite number of problems may be pro- 
posed, all similar to Prob. 5; but they are all solved by the 
formula of Prob. 6. We also see what is necessary in order 
that the answers may be obtained in whole numbers. The 
given distance (a) must be exactly divisible by m-+-n. 

Prob. 7. A gentleman meeting three poor persons, divided 
60 cents among them; to the second he gave twice, and to 
the third three times as much as to the first. What did he 
give to each? 

Let x = the sum given to the first. 

Then 2z = the sum given to the second, 
and 32 = the sum given to the third. 

Then, by the conditions, 


x+227+327=60. 
That is, 6z=60, 
or x=10. 


Therefore he gave 10, 20, and 30 cents to them respectively. 


The learner should verify this, and all the subsequent results, 
The same problem generalized. 


Prob. 8. Divide the number a into three such parts, that the 


+. 


as 


70 SIMPLE EQUATIONS. 


second may be m times, and the third 7 times as great as the 
first. 
By cipeyt GARE gee 
1+m+n’? 1+m+n’? 14m4n 
What is necessary in order that the ia values may 
be expressed in whole numbers ? | 


Ans. 


Prob. 9. A bookseller sold 10 books at a certain price, and 3 


afterward 15 more at the same rate. Now at the last sale he 
received 25 dollars more than at the first. What did he re- 
ceive for each book? 
Ans. Five dollars. 
The same Problem generalized. 
Prob. 10. Find a number such that when multiplied success- 
ively by m and by n, the difference of the products shall be a. 
Ans. jae 
m—n 
Prob. 11. A gentleman dying, bequeathed 1000 dollars to 
three servants. A was to have twice as much as B, and B 
three times as much as C. What were their respective 
shares ? : 





Ans. A received $600, B $300, and C $100. 
Prob. 12. Divide the number a into three such parts that the 
second may be m times as great as the first, and the third n 
times as great as the second. | 
A a ma mna. 
" T4m+mn? 1+m+mn? 1em+mn’ 
Prob. 13. A hogshead which held 120 gallons was filled 
with a mixture of brandy, wine, and water. There were 10 
gallons of wine more than there were of brandy, and as much 
water as both wine and brandy. What quantity was there of 
each ? 
Ans. Brandy 25 gallons, wine 35, and water 60 gallons. 
Prob. 14. Divide the number a into three such parts, that 
the second shall exceed the first by m, and the third shall be 
equal to the sum of the first and second. 
Ans. 


a—2m at2m a 

4: Ai 
Prob. 15. A person employed four workmen, to the first of 

whom he gave 2 shillings more than to the second; to the 





fever ft 


SIMPLE EQUATIONS. 71 


second 3 shillings more than to the third; and to the third 4 . 
shillings more than to the fourth. Their wages amount to 32 
shillings. What did each receive ? 
Ans. They received 12, 10, 7,and 3 shillings respectively. 
Prob. 16. Divide the number a into four such parts, that the 
second shall exceed the first by m, the third shall exceed the 
second by n, and the fourth shall exceed the third by p. 





: ay cl Geass aE? | hile ois BA 
4 4 

atM+2n—p  atm+2n+s3p 

4 : 4 : 


(107.) Problems which involve several unknown quantities 
may often be solved by the use of a single unknown letter. 
Most of the preceding examples are of this kind. In general, 
when we have given the sum or difference of two quantities, 
both of them may be expressed by means of the same letter. 
For the difference of two quantities added to the less must be 
equal to the greater; and if one of two quantities be sub- 
tracted from their sum, the remainder will be equal to the 
other. 

Prob. 17. At a certain election 36000 votes were polled ; 
and the candidate chosen wanted but 3000 of having twice as 
many votes as his opponent. How many voted for each? 

Let z = the number of votes for the unsuccessful candidate. 

Then 36000—z = the number the successful one had, 

And 36000—2z+3000=2z. 

Ans. 13000 and 23000. 

Prob. 18. Divide the number a into two such parts, that one 
part increased by b shall be equal to m times the other part. 


ma—b a+b 
m+1? m+1 

Prob. 19. A train of cars moving at the rate of 20 miles per - 
hour, had been gone three hours, when a second train followed 
at the rate of 25 miles per hour. In what time will the second 
train overtake the first ? 

Let x = the number of hours the second train is in motion, 

x+3 = the time of the first train. 
Then 25x = the number of miles traveled by the second train, 
20 (x+3)= the miles traveled by the first train. 


Ans. 








C4 “ Ei. 

2 oy 
b . x , %\ 

72 SIMPLE EQUATIONS. . 
% 7 , 


But at the time of meeting they must both have traveled the 
same distance. Re 

Therefore 25a=20z-+60. 

By transposition, 5¢=60, 
and ea 12: 

Proof. In 12 hours, at 25 miles per hour, the second train 
goes 300 miles; and in 15 hours, at’20 miles per hour, the first 
train also goes 300 miles; that is, it is overtaken’ by the sec- 
ond train. } 

Prob. 20. Two bodies move in the same direction from two 
places at a distance of a miles apart; the one at the rate of x 
miles per hour, the other pursuing at the rate of m miles per 
hour. When will they meet? 





Ans. In 


a 
hours. 
m—n 


This Problem, it will be seen, is essentially the same as 
Prob. 10. 
Prob. 21. Divide the number 197 into two such parts, that: 
four times the greater may exceed five times the less by 50. 
Ans. 82 and 115. 
Prob. 22. Divide the number a into two such parts, that m 
times the greater may exceed n times the less by b. 
ma—b na+b 
Ans. ; 
min” m+n 
When x=1, this Problem reduces to Problem 18. 
When b=0, this Problem reduces to Problem 24. 3 
Prob. 23. A prize of 2329 dollars was divided between two 
persons, A and B, whose shares were in the ratio of 5 to 12. 
What was the share of each ? 
Beginners almost invariably put z to represent one of the 
quantities sought in a problem; but a solution may often be 
very much simplified by pursuing a different method. Thus, 
in the preceding problem, we may put z to represent one fifth 
of A’s share. Then 5z will be A’s share, and 12z will be B’s, 
and we shall have the equation 
52+ 122=2329, 
and e137, 
consequently their shares were 685 and 1644 dollars. 





SIMPLE EQUATIONS. 73 


Me 

“Piob. od. Divide the number a into two such parts, that the 
first part may be to the second as m to n. 
— ma na 
‘mtn’? mtn 





Ans 


Prob. 25. What number is that whose third part exceeds its 
fourth part by 16? 

Let 122 = the number. 
Then 47 —38z=16, 
or x= 16. 
Therefore the number =12 x 16=192. 

Prob. 26. Find a number such that when it is divided suc- 
cessively by m and by n, wt ‘difievenee of the quotients shall 
be a. 


Ans. ——. 
Prob. 27. What two numbers are as 2 to 8, to each of 


which, if four be added, the sums will be as 5 to 7? 
A strict adherence to system would have required this ex- 


ample to be placed after the subject of Proportion, which is: 


treated of in Section XIII. It is, however, only necessary to 
assume one simple principle which is employed in Arithmetic, 
viz., If four quantities are proportional, the product of the ex- 
tremes is equal to the product of the means.. 


Thus, if aibvrerd. 
Then ad=be. 

In the preceding Problem, let 2x and 3z be the numbers. 
Then 97+4: 387+4::5:7, 


and by the last principle, 
147% +28=1527+-20. 

Prob. 28. What two numbers are as m to n, to each of 

which, if a be added, the sums shall be as p to q? 
Unspe Bee Big BEND OD. 
Pi elas. AG —— RP 

Prob. 29. A gentleman divides a dollar among 12 children, 
giving to some 9 cents each, and to the rest 7 cents. How 
many were there of each class ? SA 4 : 

Prob. 30. Divide the number a into two such parts, that j if 


74 SIMPLE EQUATIONS. 


the first is multiplied by m and the second by n, the sum of 
the products shall be b. 





m—n’? m—n' 
Prob. 31. If the sun moves every day one degree, and the 
moon thirteen, and the sun is now 60 degrees in advance of 
the moon, when will they be in conjunction for the first time, 
second time, and so on? 33, 68) ZS Ge , 
Prob. 32. If two bodies move in the same diipafian upon the’ 
circumference of a circle which measures a miles, the one at 
the raté of x miles per day, the other pursuing at the rate of m 
miles per day, when will they meet for the first time, second 
time, &c., supposing them to be b miles apart at starting ? 
Ans. In iss kid Ril eo &c., days. 
—n’> m—n’ m—n : 
It will be seen that this pit ae includes Prob. 20. 
Prob. 33. Divide the number 12 into two such parts, that the 
difference of their squares may be 48. 
Prob. 34. Divide the number a into two such parts, that the 
difference of their squares may be 6. 








2a.’ Bag 
Prob. 35. The estate of a bankrupt, valued at 21000 dollars, 

is to be divided among three creditors according to their re- 

_ spective claims. The debts due to A and B are as 2 to 8, 

while B’s claims and C’s are in the ratio of 4 to 5. What sum 

must each receive ? hd, YECO— 4, 74 , 
Prob. 36. Divide the number a into three parts, which ‘shall 

be to each other as m:n: p. 

Ans. oe me ae ing 
mintp m+ntp m+n+p 
When p=1, Prob. 36 1educes to the same form as Prob. 8. 
Prob. 37. A grocer has two kinds of tea, one worth 72 

cents per pound, the other 40 cents. How many pounds of 

each must be taken to form a chest of 80 pounds, which shall 
be worth 60 cents ? ; 
Ans. 50 pounds at 72 cents, and 30 pounds at 40 cents. 
Prob. 38. A grocer has two kinds of tea, one worth a cents 
per pound, the other 6 cents. How many pounds of each must 








Pi 4 


t SIMPLE EQUATIONS. 75 


be taken to form a mixture of x pounds, which shall be worth 
ec cents ? 





pounds at a cents, 


Ane. ‘bse 
a—b 
n(a—c) 
a—b 
Prob. 39. A can perform a piece of work in 6 days; B can 
perform the same work in 8 days; and C can perform the 
same work in 24 days. In what time Ay a), finish it if all 
work together ? 
Prob. 40. A can perform a piece of work in a dave B in b 
days, and C inc days. In what time will they perform it if all 
work together ? 


and pounds at 6 cents. 


Ans 





abe Lee 
‘ abtactbe 7 
Prob. 41. There are three workmen, A, B, and C. <A and 
B together can perform a piece of work in 27 days; A and C 
together in 36 days; and B and C together in 54 days. In 
what time could they finish it if all worked together ? 
A and B together can perform ,, of the work in one day. 
A and C ss En ¥ one “ 
B and C es te ic one “ 
Therefore, adding these three results, 
2A+2B+2C can PEGE a i+ s'¢+z; in one day. 
=,!, in one day. 


Therefore, A, B, and C ioastharl can perform ; of the work 
in one day; that is, they can finish it in 24 ie If we put 
x to represent the time in which they would all finish it, then 
they would together perform 4 part of the work in one day, 
and we should have 

ata teas 

Prob. 42. A and B can perform a piece of labor in a days; 
A and C together in b days; and B and C together in ¢ days. 
In what time could they finish it if all work together ? 

abe 
eats ab+ac-+be care 

This result, it will be seen. is of the same form as that of 

Problem 40. 


76 SIMPLE EQUATIONS. 


Prob. 43. A broker has two kinds of change. It takes 20 _ 


pieces of the first to make a dollar, and 4 pieces of the second 
to make the same. Now a person wishes to have 8 pieces 
for a dollar. How baad of each kind must the broker give 
him ? Ls et thd — 

Prob. 44. A has two kinds of fy de there must be a 
pieces of the first to make a dollar, and b pieces of the second 
to make the same. Now B wishes to have c pieces for a dol- 
lar. How many pieces of each kind must A give him? 








Ans. a =e of the first kind: AN si of the second. 


Prob. 45. Divide the number 45 into four such parts, that 
the first increased by 2, the second diminished by 2, the third 
multiplied by 2, and the fourth divided by 2, shall all be 
equal. 

In solving examples of this kind, several unknown quantities 
are usually introduced, but this practice is worse than super- 
fluous. ‘The four parts into which 45 is to be divided, may be 
represented thus: 


The first =r—2, 
second =2r42, 
third 2 
fourth == Das 


for if the first expression be increased by 2, the second dimin- 
ished by 2, the third multiplied by 2, and the fourth divided by 
2, the result in each case will be z. The sum of the four parts 
is 422, which must equal 45. 

Hence D=2103 

Therefore the parts are 8, 12, 5, and 20. 

Prob. 46. Divide the number a into four such parts, that 
the first increased by m, the second diminished by m, the third 
multiplied by m, and the fourth divided by m, shall all be 
equal. 

A ma ma ma 
tS (m+l1yp? ieionieeg iy (m+1)” 

Prob. 47. A merchant maintained himself for three years at 
an expense of $500 a year; and each year augmented that 
part of his stock which was not thus expended by one third 


wd 


SIMPLE EQUATIONS. V7 


thereof. At the end of the third year his orginal ier was 
doubled. What was that stock ? Ye 

Prob. 48. A merchant supported himself for heresy years at 
an expense of a dollars per year; and each year augmented 
that part of his stock which was not thus expended by one 
third thereof. At the end of the third year his original stock 
was doubled. What was that stock ? 


shat 
10 ° 


Prob. 49. A father, aad 54 years, has a son aged 9 years. 
In how many years will the age of the father be four times 


Ans. 





that of the son? D 6 oneaaw 


Prob. 50. The age of a father is Br antevented: i a, the age 
of his son by 6. In how many years will the age of the fa- 
ther be x times that of the son ? 





SECTION VIIl. 


SIMPLE EQUATIONS CONTAINING TWO 
OR MORE UNKNOWN QUANTITIES. 


(108.) In the examples which have been hitherto given, each 
problem has contained but one unknown quantity ; or, if there 
have been more, they have been so related to each other that 
all have been expressed by means of the same letter. This, 
however, can not always be done, and we are now to consider 
how equations of this kind are resolved. 

If we have two equations, with two unknown quantities, we 
must endeavor to deduce from them a single equation, con- 
taining only one unknown quantity. We must, therefore, make 
one of the unknown quantities disappear, or, as it is termed, 
we must eliminate it. There are three different methods of 
elimination which may be practiced. 

The first is by substitution, 

“ second “ comparison, 
“third “ addition and subtraction. 


ELIMINATION BY SUBSTITUTION. 


(109.) Let it be proposed to solve the system of equations 
x+ty=12 
cw—y= 6. 3 
From the second equation, we find the value of x in terms 
of y, which gives 
x=y +6. 
Substituting the expression y+6 for z in the first equation, 
it becomes 
yt6+y=12; 


SIMPLE EQUATIONS, ETC. 79 


from which we find that y=3; and since we have already 
seen that z=y+6, we find that z=3+6=9. | 

To verify these values, substitute them for z and y in the 
original equations, and we shall obtain 


9+3=12 
9—3= 6. 
Again, take the equations 


22+3y=13 
5a +4y=22. eo 
From the first equation we find 
18—2r 
age ae 
Substituting this value of y in the second equation, it becomes 
13—2zx 
ot 4X a= 2, 
an equation containing only xz, which, when solved, gives 
L==2; © 
and this value of x, substituted in either of the original equa- 
tions, gives 
y=s. 
The method thus exemplified is expressed in the following 


RULE. 


Find an expression for the value of one of the unknown quan- 
tities in one of the equations; then substitute this value in the 
place of its equal in the other equation. 


ELIMINATION BY COMPARISON. 


(110.) To illustrate this method, take equations (1.) of the 
preceding Article. Derive from each SHDELE an expression 
for x in terms of y, and we shall have 

xr=12—y, 
r= 6+yY. 

These two values of x must be equal to each other, and by 
comparing them we shall obtain 

12—y=6+y, 
an equation involving only one unknown quantity ; 
whence y=3. 


89 SIMPLE EQUATIONS 


Substituting this value of y in the expression z=6+-y, and 
we find z=9, as before. 
Again, take equations (2.) of the preceding Article. 
From equation first, we find 
, 13—2z 
oe 
and from equation second, 





4 
Putting these values of y equal to each other, we have 
13—2r 22—5z 
eS ea 
an equation containing only z, whence we obtain 
L=2. bs 
Substituting this value of z in either of the preceding ex- 
pressions for y, we find. 


*y=s. 
The method thus exemplified is expressed in the following 





RULE. 


Find an expression for the value of the same unknown quan- 
tity in each of the equations, and form a new equation by put- 
ting one of these values equal to the other. 


ELIMINATION BY ADDITION AND SUBTRACTION. 


(111.) To illustrate this method, take equations (1.) of Art. 
109. Since the coefficients of y in the two equations are 
equal and have contrary signs, we may eliminate this letter by 
adding the two equations together, whence we obtain 

22> ]8, 
or z= 9. 

We may now deduce the value of y by substituting the 
value of x in one of the original equations. Taking the first 
for example, we have 

9+y=12, 
whence Y= 

Since the coefficients of z are equal in the two original 
equations, we might have eliminated this letter by subtracting 


CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 8] 


one equation from the other. Subtracting the first from the 
second, we obtain 
2y=6, 
or y=3. 

Let us apply the same method to equations (2.) of Art. 109. 
We perceive that if we could deduce from the proposed equa- 
tions two other equations, in which the coefficients of y should 
be equal, the elimination of y might be effected by subtracting 
one of these new equations from the other. 

It is easily seen that we shall obtain two equations of the 
form required, if we multiply all the terms of each equation by 
the coefficient of y in the other. Multiplying, therefore, all 
the terms of equation first by 4, and all the terms of equation 
second by 3, they become 

8z+ 12y=8582, 
157+12y=66. 
Subtracting the former of these equations from the latter, we 
find 
iT 4. 
whence Z= 2. 

In like manner, in order to eliminate z, multiply the first of 
the proposed equations by 5, and the second by 2, they will 
become 3 

102-+15y=65, 
10z+ 8y=44. 

Subtracting the latter of these two equations from the for- 
mer, we have 

| ty=2l, 
whence y= 3. 


This last method is expressed in the following 


RULE. 


Multiply or divide the equations, if necessary, in such a man- 
ner, that one of the unknown quantities shall have the same coef- 
ficient in both. Then subtract one equation from the other, if 
the signs of these coefficients are the same, or add them together 
if the signs are different. : 

¥ 


aSer” 4 SIMPLE EQUATIONS 


EXAMPLES. 
(112.) Ez. 1. Given 5%+-4y=58 ) to find the values of z 
32+ Ty=67 and y. 
By the first method. 


From the second equation we find 


30=67—Ty. 
67—7 
Therefore r= 
Substituting this value of z in the first equation, 
67—7 
5x +-4y=58, 
Hence 335—35y-+12y=174. . 
By transposition, 835—174=35y— 12y, 
or .161=23y. 
Therefore yop 


Substituting this value of y in the expression for the value 
of x given above, it becomes 
67=7x7_67—-49_18_ 
3 3 3 3 


Thus we have y=, and z=6. 


By the second method. 


From the first equation we find 








5c=58—4y, 
58—4 
whence L= a 
5 

, 67—7 
From the second equation, z= 3 “A 

58—4y 67—7 
Therefore tae. 


Clearing of fractions, 174—12y=335—35y. 
By transposition, 35y— 12y=335—174, 
or 23y=161. 
Therefore Jerk 
whence, as before, z=6. 


+ a 
S 


CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 83 


By the third method. 
Multiplying the second equation by 5 and the first by 3, we 


obtain 
150+35y=335, 


and 152+12y=174. 
___ By subtraction, 23y=161, 
or > y= 7. 


Whence, from equation first, 
5z=58—4y=58— 28=30, 


and therefore xz=6. 


Thus the same example may be solved by either of the three 
methods, and each method has its advantages in particular 
cases. Generally, however, the first two methods give rise to 
fractional expressions which occasion inconvenience in prac- 
tice, while the third method is not liable to this objection. 
When the coefficient of one of the. unknown quantities in one 
of the equations is equal to unity, this inconvenience does not 
occur, and the method by substitution may be preferable; the 
third will, however, commonly be found most convenient. 

Ex. 2. Given 11z+3y=100 

4x—Ty= 4 

Multiplying the first equation by 7 and the second by 3, we 

obtain 


to find the values.of x and y. 


171x+21y=700, 


12z—2ly= 12. 
Therefore, by addition, 897=712, 
or z= 8. 
From equation first, 3y=100—11z, 
=100—88=12, 
and y=4, 


These values of x and y may be easily verified by substitu- 
tion in the original equations. 
Thus, 11X8+3xX4=100; or 88+12=100. 


And 4X8—7X4= 4; or 32—28= 4. 
Ez. 3. Given mal 
to find the values of x and y. 
Ye aig i 
PP ey 


Ans. x=6, y=12. 


84 SIMPLE EQUATIONS 


Ex. 4. Given + 8y=31 


’ 5 to find the values 2 x wie y 





4" 4+107=192 
E+ Se Sandy | 
Ex. 5. Given 2y oa ee +t to find the values of 
8— Q7+1 x and y. 
Se pes == 24} — 2 ioe 


m/ 


b Sind, i) § - : 
Ex. 6. Given “4° <m| 
; f to find the values of x and y. 
—+-=n 
y 
D PC aed, bc— Liat ue be—ad 


nb—m me—na 


(113.) When a problem involves a large number of quanti- 
ties, it is common to designate a part of them by different let- 
ters, and for the remaining quantities to employ the same let- 
ters accented or numbered. 


Thus, a, a’, a”, a’, al” : : ; ; . am 
a”, a®, a®, a® : e : ; - a” 
athe, 1 ¥Gee...\ he : - : ; ‘Seite 
ay, Diy Qiiys Qy 41 be +d am 


are used to denote different quantities, oars they cere 
imply some connection between the quantities which they rep- 
resent. a’ is read a prime; a", a second; a’, a third, &c. 
We must carefully distinguish between a, and a’; between a, 
and a‘, &c. In the one case, the numerals are exponents, and 
denote powers of a; while in the other case, the numerals are 
only used for the sake of convenience to denote distinct quan- 
tities. Examples showing the convenience of this notation will 


be found in Sections XIX. and XX. 
.7. Gi by = . 
Ex iven seine 2 ; to find the values of x and y, 


b’c—be' ac'—a'c 
ss e ab —a'b? 7~ ab'—a'b 
The symmetry of these expressions is well calculated to fix 
them in the memory. 


Ex. 8. What fraction is that, to the numerator of which, if 4 


CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 85 


be added, the value is one half; but if 7 be added to the de- 
nominator, its value is one fifth ? 


x . 
Let * represent the fraction required. 


Then, by the first condition, 


t+4 ar 
aE err mt whence 22+8= =, 
By the second condition, 
#7; whence 5¢=y+7 
yt? whence 5e=y+7. 
Subtracting the first equation from the second, we have 
whence 327=15, 
or r=5d. 
Therefore, y=2r+8=1048=18, 
and the fraction is a. 
~“ 5+4 1 
Proof. aa aw 
5 1 
ane iet7-5 


Ex. 9. A certain sum of money, put out at simple interest, 
amounts in 8 months to $1488, and in 15 months it amounts to 


$1530. What is the sum and rate per cent. ? Grr )4 Yet 


Ez. 10. A sum of money put out at, simple interest amounts ® /7.° 


in m months to a dollars, and in m months to b dollars. 
Required the sum and rate per cent. ? 





; the rate is 1200 





_ na—m 
Ans. The sum is 
_ na—mb 


Ex. 11. There is a number consisting of two digits, the 
second of which is greater than the first; and if the number 
be divided by the sum of its digits, the quotient is 4; but if 
the digits be inverted, and that number be divided by a num- 
ber greater by two than the difference of the digits, the quo- 
tient is 14. 

Required the number. 

Let x represent the left hand digit, 
and y 3 right hand digit. 

Then, since x stands in the place of tens, the number will be 
represented by 10z+y. 


86 SIMPLE EQUATIONS 


Hence, by the first condition, 





l0r+y _ 
arty 
By the second condition, 
10y+z. 
y—at2 
Whence x=4, y=8, 


and the required number is 48. 

Ex. 12. A boy expends thirty pence in apples and pears, 
buying his apples at 4 and his pears at 5 for a penny, and 
afterward accommodates his friend with half his apples and 
one third of his beats for 13 spon How bie a he buy 
of each ? MMe box. Va 

Ex. 13. A father ARs a sum “of money to be divided among 
his children, as follows: the first is to receive $300. and the 
sixth part of the remainder; the second $600 and the sixth 
part of the remainder; and, generally, each succeeding one 
receives $300 more than the one immediately preceding, to- 
gether with the sixth part of what remains. At last it is found 
that all the children receive the same sum. What was the 
fortune left and the number of children ? 

Ans. The fortune was $7500, the number of children 5. 

Fiz. 14. A sum of money is to be divided among several 
persons, as follows: the first receives a dollars together with 
the nth part of the remainder; the second 2a together with 
the nth part of the remainder; and each succeeding one a dol- 
lars more than the preceding, together with the nth part of 
the remainder; and _ it is found, at last, that all have received 
- the same sum. What was the amount divided, and the num- 
ber of persons ? 

Ans. The amount =a(n—1)’, the number of persons =n—1. 


EQUATIONS WHICH CONTAIN THREE OR MORE UNKNOWN 
QUANTITIES. 


_(114.) Let us now consider the case of three equations in- 
volving three unknown quantities, 
Take the system of equations, 
3r+2y+ z=16, (1.) 
Qa -+2y+22=18, (2.) 
2e+3y+ 2=17. (3.) 


CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 87 


In order to eliminate z between equations (1.) and (2.), we 
will divide both members of the second equation by two; we 
thus obtain 

r+yt+z=9. 

Subtracting this from the first equation, we find a new equa- . 
tion containing but two unknown quantities, 

F Qa y=7. (a.) 

In order to eliminate z between equations (1.) and (3.), sub- 
tract the former from the latter, which gives 

—z+y=1. (G.) 

From the two equations (a.) and (G.), one may be deduced 
containing only one unknown quantity. For, by subtracting 
the one from the other, we have 

SE =G. OF 2s 

Substituting this value of z in equation (G.), we obtain 

y=. 

Substituting these values of z and y in equation (1.), we ob- 
tain | 

3X2+2X3+2=16., 

Hence gm, 

These values of x, y, and z may be verified by substitution 
in the original equations. | 

We have effected the elimination in this case by method 
third, Art. 111; but either of the other methods might have 
been employed. Hence, to solve three equations containing 
three unknown quantities, we have the following 


RULE. 


(115.) From the three equations, deduce two containing only 
two unknown quantities; then from these two deduce one con- 
taining only one unknown quantity. 

Ez. 15. Given z+ y+ 2=29 (1) 

x+2y+3z=62 (2.) > to find a, y, and z. 
17+-1y+12=10-(3.) 

Subtract equation (1.) from (2.), and we obtain 

y+2z2=38 ; (a.) 
clearing equation (3.) of fractions, we have : 
62+4y+32=120. (4.) 


88 SIMPLE EQUATIONS 


~ Multiplying equation (1.) by 6, 

r+ 6y-+62=174. (5.) 
Subtracting (4.) from (5.), 2y+3z=54. (G.) 
We have thus obtained two equations, (2). and (@.), contain- 

- ng two unknown quantities. 

Multiplying (a.) by 2, we have 2y+4z=66. (6.) 
Subtracting (@.) from (6.), z=12. 
Substituting this value of z in (@.), we obtain 


2y+36= 54. 
Whence y==9. 
Substituting these values of y and z in equation (1.), 
L+9+12=29. 
Whence cay 


These values may be verified as in former examples. 
Ex. 16. Given 2x+4y—3z=22 
42 —2y+5z=18 c find x, y, and z. 
6z2+Ty— z2=63 
Ans. c=3, y=7, 2=4. 


Ex. 17. Given x+y=a ve ae po 
z+z=b c fiid +49) ind z. "ype p 
y+z=c ; = 


Ez. 18. Given z+1y+12=82 f 
tea+iytiz=15 c find x, y, and z. 
pet+tyt gz=12 

(116.) If we had four equations involving four unknown 

quantities, we might, by the methods already explained, elim- 
inate one of the unknown quantities. We should thus obtain 
three equations between three unknown quantities, which might 
be solved according to Art. 114. So, also, if we had five 
equations involving five unknown quantities, we might, by the 
same process, scat them to four equations RE pee four 


unknown quantities; then to three, and so on. By following 
the same method, we might resolve a system of any number 


of equations of the first degree. Hence, if we have m equa- 
tions involving m unknown quantities, we proceed by the fol- 
lowing 


RULE. 


1. Combine successively any one of the equations with each 
of the others, so as to eliminate the same unknown quantity ; we 


7 


CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 89 


thus obtain m—1 new equations containing m—1 unknown 
quantities. 

2. Kliminate another unknown quantity by combining any 
one of these new equations with the others; there will result 
m—2 equations containing m—2 unknown quantities. 

3. Continue this series of operations until there results a 
single equation containing but one unknown quantity, from 
which the value of this unknown quantity is easily deduced. 
Then by going back, step by step, to one of the original equa- 
tions, the values of the other unknown quantities may be suc- 
cessively determined. 


Ez. 19. Given 7x—2z +3u=17 
4y—2z+ t=11 | 
5y—32—2u= 8 > to find z, y, z, u, and z. 
4y—3u+2t= 9 
3z+8u=33 J 
ANS. ten too yo aw, toe 
Either of the unknown quantities may be selected as the 


one to be first exterminated. It is, however, generally best to 
begin with that which has the smallest coefficients ; and if each 


of the unknown quantities is not contained in all the proposed 


equations, it is generally best to begin with that which is found 
in the least number of equations. 
Ex. 20. A person owes a certain sum to two creditors. At 


one time he pays them $530, giving to one four elevenths of - 


the sum which is due, and to the other $30 more than one 
sixth of his debt to him. Ata second time he pays them $420, 
giving to the first three sevenths of what remains due to him, 
and to the other one third of Benet remains vide to him. Wobatt 
were the debts ? M 

Ez. 21. If A and B paethent can per Soren a piece ip Ok 
in 12 days, A and C ipsther in 15 days, and B and C in 20 
days, how many days will it take each Pes son to aa form the 


a 


same work alone? Jt, DtdleveiwW 


This Problem is readily solved by first a in seats time 


they could finish it if all worked together. 
Ex. 22. If A and B together can perform a piece of work 
in a days, A and C together in 6 days, and B and C inc days, 


» 


90 SIMPLE EQUATIONS 


how many days will it take each person to perform the same 
work alone ? 


2Qabe 

Ans. A requires —— ncn aa days, 
2abe 

e ab+be—ac gaye 
2abe 

° ab-+-ac—be days, 


(117.) Hitherto we have supposed the number of equations 
equal to the number of symbols employed to denote the un- 
known quantities. This must be the case with every problem, 
in order that it may be determinate; that is, that it may not 
admit of an indefinite number of solutions. 

Suppose, for example, that a problem involving two un- 
known quantities (2 and y) leads to the single equation 

L—Y=3. 
Now if we make y=1, then 7=4; 
y=2, then r=5; 
y=3, then 7=6 ; 
w= 4 then a7, 
&e., cus 
and each of these systems of values, 1 and 4, 2 and 5, 3 and 6, 
&c., substituted for 2 and y in the original equation, will sat- 
isfy it equally well. Hence the problem is indeterminate ; that 
is, admits of an indefinite number of solutions. 

(118.) If we had two equations involving three unknown 
quantities, we could, in the first place, eliminate one of the un- 
known quantities by means of the proposed equations, and 
thus obtain one equation containing two unknown quantities, 
which would be satisfied by an infinite number of systems of 
values. Therefore, in order that a problem may be determ- 
inate, its enunciation must contain as many different condi- 
tions as there are unknown quantities, and each of these con- 
ditions must be expressed by an independent equation. 

Equations are said to be independent when they express 
conditions essentially different ; and dependent when they ex- 
press the same conditions under different forms. 

Thus, z+y= 7 


Qe-+-y=10 are independent equations, 


CONTAINING TWO OR MORE UNKNOWN QUANTITIES. 91 


But z+ y= 7 
2x4+2y=14 
because the one may be deduced from the other. 

(119.) If, on the contrary, the number of independent equa- 
tions exceeds the number of unknown quantities, these equa- 
tions will be contradictory. 

For example, let it be required to find two numbers such 
that their sum shall be 7, their difference 1, and their product 
100. 

From these conditions we derive the following equations : 

xty= 7, 
xz—y= 1, 
xy=100. 

From the first two equations we easily find 

x=4, and y=3. 

Hence the third condition, which requires that their product 
shall be equal to 100, can not be fulfilled. 


' are not independent, 


SECTION IX. 


DISCUSSION OF EQUATIONS OF THE 
FIRST DEGREE. INEQUALITIES. 


(120.) To discuss a problem or an equation is to determine 
the values which the unknown quantities assume for particular 
hypotheses made upon the values of the given quantities, and 
to interpret the peculiar results obtained. The term, there- 
fore, is not strictly applicable, except to problems which are 
stated in the most general form, like some of those in Arts. 106 
and 107. If the sum of two numbers is represented by a and 
aa difference by b, the greater number will be expressed by 
a ate and the less by faa Here a and b may have any 
values whatever, and os nee formule will always hold true. 
It freqently happens that, by attributing different values to the 
letters which represent known quantities, the values of the un- 
known quantities assume peculiar forms which deserve con- 
sideration. 

(121.) We may obtain five species of values for the unknown 
quantity in a problem of the first degree. 

I. Positive values. 

II. Negative values. 


III. Values of the form of zero, or 7 rr vw 


IV. Values of the form of a OD = | ; ie, 


V. Values of the form of - 


We will consider these five cases in succession. 


DISCUSSION OF EQUATIONS, ETC. 93 


I. Positive values are generally answers to problems in the 
sense in which they are proposed. Nevertheless, all positive 
values will not always satisfy the enunciation of a problem. 
‘If, for example, a problem requires an answer in whole num- 
bers, and we obtain a fractional value, the problem is impossi- 
ble. Thus, in Problem 17, page 71, it is implied that the value 
of x must be a whole number, although this condition is not 
expressed in the equations. It would be easy to change the 
data of the problem so as to obtain a fractional value of =z, 
which would indicate an impossibility in the problem pro- 
‘posed. Problem 48, page 76, is of the same kind; also Ez. 
11, page 85. 

If the value obtained for the unknown quantity, even when 
positive, does not satisfy all the conditions of the problem, the 
problem is impossible in the form proposed. | 

(122.) Il. Negative values. 

Let it be proposed to find a number, which, added to the 
number 0, gives for a sum the number a. 

Let z = the required number. 

Then, by the terms of the problem, 

«+b=a, whence z=a—b. 

This formula will give the value of z for every case of the 

proposed problem. 


For example, let i ald. 4: 
Then z=1—4=38. 
Again, let a=5, and b=8. 
Then xr=5—8=—83. 


We thus obtain for z a negative value. How is it to be in- 
terpreted 2 _ | 
* By referring to the problem, we see that it is proposed to 
find a number which, added to 8, shall make it equal to 5. 
Considered arithmetically, the problem is plainly impossible. 
Nevertheless, if in the equation 8+z=5, we substitute for +z 
its value —3, it becomes 


8—3=—5, 
an identical equation; that is, 8 diminished by 8 is equal to 5. 
The negative solution z=—3, shows, therefore, the impossi- 


bility of satisfying the enunciation of the problem as above 
stated ; but, taking this value of z with a contrary sign, we see 
that it satisfies the enunciation when modified as follows : 


94 DISCUSSION OF EQUATIONS 


To find a number which, subtracted from 8, gives a differ- 
ence of 5; an enunciation which differs from the former only 
in this, that we put subtract for add, and difference for sum. 

If we wish to solve this new question directly, we shall 
have 

8—zr=5. 

Whence x=8—5, or x=3. 

(123.) For another example, take Problem 50, page 77. 
The age of the father being represented by a, and that of the 

a—nb 
son by b; then any 


fore the age of the father will be n times that of the son. 


will represent the number of years be- 


Thus, suppose a=—54, b=9, and n=4. 
_ 54—36 -18 _ 

That is, the father having lived 54 years and the son 9, in 6 
years more the father will be 60 years old and the son 15. 
But 60 is 4 times 15; hence this value, z=6, satisfies the enun- 
ciation of the problem. 

Again, suppose a=45, b=15, and n=4. 

45-60 —15__ 
Te 8: eee Wma 

Here again we obtain a negative solution. How are we to 
interpret it ? | 

By referring to the problem, we see that the age of the son 
is already more than one fourth that of the father, so that the 
time required is already past by five years. The value of x 
just obtained, taken with a contrary sign, satisfies the following 
enunciation : | 

A father is 45 years old, his son 15; how many years since 
the age of the father was four times that of his son? 

The equation corresponding to this new enunciation is 

45—x 
4 

Whence 60—4¢2=45—z; and z=5. 

(124.) Reasoning from analogy, we deduce the following 
general principles: : 

1. Every negative value found for the unknown quantity in a 


Then 


Then 





15-—z2= 





OF THE FIRST DEGREE. 95 


problem of the first degree, indicates an absurdity in the condi- 
tions of the problem, or at least in its algebraic statement. 

2. This value, taken with a contrary sign, may be regarded 
as the answer to a problem, whose enunciation only differs from 
that of the proposed problem in this, that certain quantities 
which were avpep should have been sustrracren, and recipro- 
cally. ; 

(125.) In what case would the value of the unknown quan- 
tity in Prob. 20, page 72, be negative ? | 

“ Ans. When n>m. 
Thus, let m=20, n=25, and a=60 miles. 


60 60 
Then Oe nOR pc. H ie 


To interpret this result, observe that it is impossible that the 
second train, which moves the slowest, should overtake the 
first. At the time of starting, the distance between them was 
60 miles, and every subsequent hour the distance increases. 
If, however, we suppose the two trains to have been moving 
uniformly along an endless road, it is. obvious that at some 
former time they must have been together. 

This negative solution then shows an absurdity in. the con- 
ditions of the problem. The problem should have been stated 
thus : 

Two trains of cars, 60 miles apart, are moving in the same 
direction, the forward one 25 miles per hour, the other 20. 
How long since they were together ? 

To solve this problem, let 2 = the required number of hours. 

Then 25x = the distance traveled by the first train, 





—12. 


200 = . i: second train. 
And since they are now 60 miles apart, 
25x—=202+ 60. ‘ 
Hence 52=60, 
and jy) w=+12. 


We thus obtain a positive value of z. 

In order to include both of these cases in the same enuncia- 
tion, the question should have been asked, Required the time of 
their being together, leaving it uncertain whether the time was 
past or future. 

In what case would the value of one of the unknown quan 


96 DISCUSSION OF EQUATIONS 


tities in Problem 84, page 74, be negative? Why should it be 
negative? and how could the enunciation be corrected for this 
case ? 18", 

In what case would the value of one of the unknown quan- 


tities in Problem 4, page 67, be negative? ~» A @ o> 


(126.) III. Values of the form of zero, or 7 

In what case would the value of the unknown quantity in 
Problem 20, page 72, become zero, and what would, this value 
signify ? 

Ans. This value becomes zero when a=0, which signifies 
that the two trains are together at the outset. 

In what case would the value of the unknown quantity in 
Problem 50, page 77, become zero, and what would this value 
signify ? 

Ans. When a=nb, which signifies that the age of the fa- 
ther is now n times that of the son. 


In what case would the values of the unknown quantities in__* 
Problem 38, page 75, become zero, and what would this sig- » /_ 


nify ? y . Vy 3 
When a problem gives zero for the value of the unknown 
quantity, this value is sometimes applicable to the problem, 
and sometimes it indicates an impossibility in the proposed 
question. 


(127.) IV. Values of the form of a 


In what case does the value of the unknown quantity in 
| aga . 
Problem 20, page 72, reduce to a and how shall we inter- 


pret this result ? 
Ans. When: m=n. 


On referring to the enunciation of the problem, we see that 
it is absolutely impossible to satisfy it ; that is, there can be 
no point of meeting, for the two trains being separated by the 
distance a, and moving equally fast, will always continue at 


’ a 
the same distance from each other. The result 9 may then 


be regarded as indicating an zmpossibility. 


OF THE FIRST DEGREE. 97 


a. ebetcls 
The symbol 9 18 sometimes employed to represent infinity ; 


and for the following reason : 
When the difference m—x, without being absolutely nothing, 





is very small, the quotient is very large. 


m—n 
For example, let m—n=0.01. 
Then ee Ty. 
} m-=n Ol 
Let m—n=0.0001, 
Dee RNS 
yew GOT 10000a. 


Hence, if the difference in the rates of motion is not zero, 
the two trains must meet, and the time will become greater 
and greater as this difference is diminished. Jf, then, we sup- 
pose this difference less than any assignable quantity, the time 

a 





represented by will be greater than any assignable quan- 


m—n 
lity, or infinite. 


A 
Hence we infer, that every expression of the form ri. found 


for the unknown quantity, indicates the impossibility of satis- 
fying the problem, at least in finite numbers. 
In what case would the value of the unknown quantity in 


Problem 10, page 70, reduce to the form 4 and how: shall 


Lad 


we interpret this result ? 2: 
0 
(128.) V. Values of the form of 7 


In what case does the value of the unknown quantity in 


4 


0 
Problem 20, page 72, reduce to 0! and how shall we interpret 


this result ? 
Ans. When a=0, and m=n. 
To interpret this result, let us recur to the enunciation, and 
observe that, since a is zero, both trains start from the same 
point; and since they both travel at the same rate, they will 
always remain together, and therefore the required point of 
meeting will be any where in the road traveled over. The 


G 


Sse Me 
98 OF ZERO AND INFINITY. 
problem, then, is entirely indeterminate, or admits of an infinite 
number of solutions, and the expression : may represent any 
finite quantity. 

We infer, therefore, that an expression of the form : found 


for the unknown quantity, generally indicates that it may have 
any value whatever. In some cases, however, this value is 
subject to limitations. 

In what case would the values of the unknown quantities in 


0 
Problem 44, page 76, reduce to 5. and how would they satisfy 


the conditions of the problem ? ; 
Ans. When a=b=c, 
which indicates that the coins are all of the same value. B 


might therefore be paid in either kind of coin; but there is a | 


limitation, viz., that the value of the coins must be one dollar. 
In what case do the values of the unknown quantities in 


0 : 
Problem 38, page 75, reduce to 0: and how shall we interpret 


this result? br ve se 


i —— 4 sf 


OF ZERO AND INFINITY. 


(129.) From Art. 127, it is seen that in Algebra we some- 
times have occasion to consider infinite quantities. It is nec- 
essary, therefore, to establish some general principles respect- 
ing them. | 

An infinite quantity is one which exceeds any assignable limit. 
It is often expressed by the character ©. Thus, a line pro- 
duced beyond any assignable limit is said to be of infinite 
length. A surface indefinitely extended, and also a solid of 
indefinite extent in any one of its three dimensions, ‘are ex- 
amples of infinity. , 

An infinite quantity does not mean an infinite number of 
terms. Thus, the fraction 1} reduced to a decimal, is .333333, 
&c., without end, but the value of this series is less than 
unity. 

Infinite quantities are not all equal among themselves. 


ee 


OF ZERO AND INFINITY. 99 


Thus the series 1+1+1+1414-4, &c., 
2424242424, &., 
3+34+34+3+3-+, &c., 
continued to an infinite number of terms, will each be infinite, 
although the second series will be double, and the third treble 
the first. 

So, also, a line may be infinitely extended both ways; or it 
may be infinitely extended in one direction, and limited in the 
other. In either case, the line is said to be infinite. 

A quantity less than any assignable quantity is called an in- 
finitesimal, and is sometimes represented by 0. 

Thus, take the series-of fractions 55/535. r355: tobs7) LC: 
By increasing the denominator, we diminish the value of the 
fraction ; and if the denominator be made infinitely great, the 
quotient will be infinitely small. 


' a 
(130.) We have seen, in Art. 127, that re where a may 
represent any finite quantity. That is, 
If a finite quantity be divided by zero, the quotient is infinite. 
a 
From the same equation we deduce =F. That is, 


If a finite quantity be divided by infinity, the quotient is zero. 
From the same equation we deduce a=0X@. That is, 
If zero be multiplied by infinity, the product is a finite quan- 


tity. 


If a finite quantity be multiplied by a proper fraction, it will 
be diminished, and the smaller the multiplier, the less the prod- 
uct. Hence, if the multiplier be infinitely small, the product 
will be infinitely small, or ax0=0. That is, 

If a finite quantity be multiplied by zero, the product will be 
er. 

From this equation we deduce a=e that is, 

If zero be divided by zero, the quotient may be any finite 
quantity. | 

The greater the multiplier, the greater will be the product. 
Hence, if a finite quantity be multiplied by infinity, the product 
will be infinite ; that is, 

aXD=O@. 


eT 


100 OF INEQUALITIES. 


F : @ d 
From this equation we deduce a at that is, 


If infinity be divided by infinity, the quotient may be any 
finite quantity. 

An infinite quantity can not be increased by the addition of 
a finite quantity, or diminished by its subtraction; that is, 
u-niaan 

So, also, a finite quantity is not altered by the addition or 
subtraction of zero; that is, ak0=a. 


OF INEQUALITIES. 


(131.) In discussing algebraical problems, as shown in Arts. 
120-128, it is frequently necessary to employ inequalities, or 
expressions of two quantities which are not equal to each oth- 
er. Generally, the principles already established for the trans- 
formation of equations are applicable to imequalities also. 
There are, however, some important exceptions to be noted, 
arising chiefly from the use of negative expressions as quan- 
tities. 

Two inequalities are said to subsist in the same sense when 
the greater quantity stands at the left in both, or at the right 
in both; and in a contrary sense when the greater quantity 
stands at the right in one, and at the left in the other. 

Thus, 9>7 and 7>6. 

As also 5<8 and 3<4, | 
are inequalities which subsist in the same sense; but the ine- 
qualities 

10>6 and 8<7, 
subsist in a contrary sense. 

(182.) I. If we add the same quantity to both members of an 
inequality, or subtract the same quantity from both members, the 
resulting inequality will always subsist in the same sense. 

Thus, 8>8. 

Adding 5 to each member, 

8+5>3+5; 
and subtracting 5 from each member, 
S+5>3—5, 

Again, take the inequality —8<—2. 


OF INEQUALITIES. 101 


Adding 6 to each member, we have 
—346< —246, or 3<4; 
and subtracting 6 from each member, 
—3— 62 —9e-6,.0r.—-9< —8. 

The student must here bear in mind what was stated in Art. 
47, of two negative quantities, that is the /east whose numer- 
ical value is the greatest. 

This principle enables us to transpose any term from one 
member of an inequality to the other by changing its sign. 


Thus, a+b’>3bh?—2a’. 
Adding 2a’ to each member of the inequality, it becomes 
a’+b’?+2a’> 3b’. 


Subtracting b’ from each member, 
a’+2a’>3b’?—0’, 
or 3a’ > 2b". 
_(183.) IL. Df we add together the corresponding members of 
two or more inequalities which subsist in the same sense, the re- 
sulting inequality will always subsist in the same sense. 


Thus, ~ 5>4 
4>2 
7>3 

Adding, we obtain 16>9. 


Il]. But if we subtract the corresponding members of two or 
more inequalities which subsist in the same sense, the resulting 
inequality will Nor ALWAyYs subsist in the same sense. 

Take the inequalities 4<17 

| 2<3 
Subtracting, we have 4—2<7—8, or 2<4, 
where the resulting inequality subsists in the same sense. 

But take 9<10 
and 6< 8. 

Subtracting, the result is 9—6> (not <) 10—8, or 3>2, 
where the resulting inequality subsists in the contrary sense. 

We should therefore avoid as much as possible the use of 
this transformation, or when we employ it, determine in what 
sense the resulting inequality subsists. 

(134.) IV. If we multiply or divide the two members of an in- 
equality by a positive number, the resulting inequality will sub- 
‘sist in the same sense. 


102 OF INEQUALITIES. 


Thus, if ax b. 
Then ma<mb, 

a b 
And ra Ger 
Also, if —a>—b. 
Then —na>—nb. . 
And | voy a 

n n 


This principle will enable us to clear an inequality of frac- 

tions. Thus, suppose we have 
a—b". oe’ —a 
a Tg 

Multiplying both members by 6ad, it becomes 

3a(a’*—b*)>2d(c’—d’). 

V. If we multiply or divide the two members of an inequality 
by a negative number, the resulting inequality will subsist in a 
contrary sense. 

Take, for example, 8>7. 

Multiplying both members by —3, we have the opposite in- 
equality, 


ne Et ot AE 
So, also, 15>12. 
Dividing each member by —3, we have 
—5<~—4, 


Therefore, if we multiply or divide the two members of an 
inequality by an algebraic quantity, it is necessary to ascer- 
tain whether the multiplier or divisor is negative, for in this 
case the inequality subsists in a contrary sense. 

VI. If we change the signs of both members of an inequality, 
we must reverse the sense of the inequality, for this transforma- 
tion is evidently the same as multiplying both members by 
tale . 

(135.) VII. Lf both members of an inequality are positive 
numbers, we can raise them to any power without changing the 
sense of the inequality. 

Thus, 5>3, 
so also, 5 >-3", or 25>>0. 


OF INEQUALITIES. 108 


And if a >b, 
then will a”">b". 

VIII. Lf both members of an inequality are not positive num- 
bers, and they be raised to any power, the resulting inequality 
will not always subsist in the same sense. 


Thus, ee, 
gives (—2)’<3’, or 4<9, 
where the resulting inequality subsists in the same sense. 
gives (—3)’<(—5)’, or 9< 25, 


where the resulting inequality subsists in a contrary sense. 

IX. In extracting the root of both members of an inequali- 
ty, it is sometimes necessary to reverse the sense of the ine- 
quality. 


Thus, from 9< 25, 
by extracting the square root, we obtain 
either 3< 5, 
or —3>-—5. 
EXAMPLES. 


1. Given 77—38< 25, to find the limit of z. 
Ans. 7<4. 


2. Given 2n-+5—-8<6, to find the limit of z. 
Ans. x<6. 


3. Given LS SAR DML to find the limit of z. 


4. Given ber acc | 
eae, >; to find the limits of z. 
maaetae | 


_ 5. A man being asked how many dollars he gave for his 
watch, replied, If you multiply the price by 4, and to the 
product add 60, the sum will exceed 256; but if you multiply 
the price by 3, and from the product subtract 40, the re- 


104 OF INEQUALITIES. 


mainder will be Beh than a Baee a Pree of the 
watch. : att 

6. What number is that sihoss half at third shart adden 
together are less than 105, but its half dimes by its at 
part is greater than 33? oi Ly a ea PX. XPS ls 

7. The double of a number diminished by 6 is greater mE 
24, and triple the number diminished by 6 is less than double 
the number increased by 10. Required the number. 


« a re) 
Y< Ae) - 
a ~~ 


SECTION X. 


INVOLUTION AND POWERS. 


(136.) According to Art. 20, the products formed by the suc- 
cessive multiplication of the same number by itself are called the 
powers of that number. 

Thus, the first power of 3 is 3. 

The second power of 3 is 9, or 3X3. 

The fourth power of 3 is 81, or 3X3X3xX3, 

&ec., &ec., &c. 

According to Art. 21, the exponent is a number or letter writ- 
ten a little above a quantity to the right, and denotes the number 
of times that quantity enters as a factor into a product. 

Thus, the first power of a is a’, where the exponent is 1, 
which, however, is commonly omitted. 

The second power of a is aXa, or a’, where the exponent 2 
denotes that a is taken twice as a factor to produce the pow- 
er aa. 

The third power of a is aXaXa, or a’, where the exponent 
3 denotes that a is taken three times as a factor to produce 
the power aaa. 

The fourth power of a is aXaXaxXa, or a’. 

Also, the nth power of a isaXaxXaxXa . . . repeated 
as a factor 7 times, and is written a”. 

Eixponents may be applied to polynomials as well as to mo- 
nomials. 

Thus (a+b-+c)* is the same as 

(a+b+c) x (a+b+c) x (a+b+c), 
or the third power of the entire expression a+b-+c. 

(137.) According to the rule for the multiplication of mono- 

mials, Arts. 49 and 50. 


106 INVOLUTION AND POWERS. 


(3ab*)’=3ab’ x 3ab’?=9a"b*. 
So, also, (4a*bc*)’ =4a°bc? x 4a°bc' = 16a'b*c’. 
Hence it appears that, 7m order to square a monomial, we must 
square its coefficient, and multiply the exponent Tee each of the 
letters by 2. 


EXAMPLES. 


1. Required the square of Tazy. 
ais 40e'e rey 
2. Required the square of 1la*bcd’. 3 / 
3. Required the square of l2a°xzy. = | * “#7 te ie 
4. Required the square of 15ab’cz*. ee aa 


9 
7 1, & 4 
x wy 


5. Required the square of 182*yz°. mg Aw 
According to Art. 53, + multiplied by +, and — 5 atileplign by. 
—,give +. Now the square of any quantity being the product 

of that quantity by itself, it necessarily follows that whatever 
may be the sign of a monomial, its square must be affected with 
the sign +. 

Thus the square of +38az or of —8az is + 9a°x’ 

(138.) The method of involving a quantity to any power, is 
easily derived from the preceding principles. 

Let it be required to form the fifth power of 2a°b’. 

According to the rules for multiplication, 

(2a°b*)°=2a°b’ x 2a°b* X 2a*b’ X 2a°b’ x 2a*b® 
= 920". 

Where we perceive 

1. That the coefficient has been raised to the fifth power. 

2. That the exponent of each of the letters has been multi- 
plied. by 5 

In like manner, 

(3a°b*c)’=8a'b'c X 3a°b'c X 3a’bic 
=3°¢ B-f-2-4-8 78-18-45 7s 1-1-f-1 
= 27a Be" 

Hence, to raise a monomial to any power, we have the fol- 

lowing 


RULE. 


Raise the numerical coefficient to the given power, and multi- 
ply the exponent of each of the letters by the exponent of the 
power required. 


INVOLUTION AND POWERS. 107 


EXAMPLES. 
. Required the fourth power of 4ab’c’. 


Neel 


Ans. 256a‘b*c”, 
. Required the fifth power of 8az‘y’. md LYS at xis 
. Required the third power of 6ry7’z*. BIO ay” 
. Required the sixth power of 2ad’y’v. Bes 0d Ay 28 
. Required the seventh power of © hag 1  YhE 6 
. Required the sixth power of 5w*xy’z SB25 3 
(139.) Let us now consider the sign with mitch the power 
should be affected. 7 
We have seen, Art. 187, that whatever may be the sign of a 
monomial, its square is always positive. It is obvious, from the 
same considerations, that the product of an even number of 
negative factors is positive, but the product of an odd number 
of negative factors is negative. 
Thus, —aX—a=+a 
—ax—ax—a=—a’ 
—aX—aX—axX—a=+e 
—aX—aX—aX—ax—-—a=—-@ 
&c., &c., &c. 
The product of several factors which are all positive, is in- 
variably positive. Hence, 
Every even power is positive, but an opp power has the same 
sign as its root. 


oO Ol Rm CO 0 


EXAMPLES. 


1. Required the square of —2z*. 


2. Required the square of —3z”. L, 

3. Required the cube of —3a’. 

4. Required the fourth power of —8a°bh. 

5. Required the fifth power of —2a’x 3z’y. , —$~R/ 

(140.) A fraction is involved by involving both the numerator 
and denominator. 


1. Thus, the square of ; is er which, by Art. 89, is 


equal to a which, by Art. 68, may be written a*b—. 








108 INVOLUTION AND POWERS. 
2ab’ 
32.6 
Ans 37 oe a’b’c— 
ab ae 
3. Required the nth power of ar, COP ae aa 


(141.) Hence, expressions with negative exponents are in- 
volved by the same rule as those with positive exponents. 
Thus, let it be required to find the square of a— 


{ : PAF ATY ; . 
This expression may be written = which, raised to the 


1 
second power, becomes a or a—, the same result as would be 


obtained by multiplying the exponent — 3 DY 2, 
fix. 1. Required the square of 83a°O-*., Ft 7 7 oe 
Ex. 2. Required the square of Tab’cde. 7 * a eae 
Ex. 3. Required the cube of —6ab—dy~’. sialik 
Ez. 4. Required the fourth power of 3a—b. 

Ez. 5. Required the fifth power of —Zab—*c’. 
(142.) A polynomial is involved by multiplying it into itself 
as many times less one as is denoted by the exponent of the 
power. 
Ez. 1. Required the fourth power of a+0. 
a+b 
a +b 
a’+ab 
+ab+b? 
(a+b)’=a?+2ab+b*, the second power of a+b. 
a+b 
a’+2a’*b+ab’ 
+ a’b+2ab’+b 
(a+b)*=a'+3a’b+3ab’+0*, the third power. 
a +b ‘ 
a‘+3a°b+3a°b’+ ab* 
+ a°b+3a7b’+3ab'+b* 
(a+b)'=a'+4a'b+6a'b'+4ab*+b*, the fourth power. 
Ex. 2. Required the fourth power of a—b. 
Ans. a‘—4a°b+6a’b’—4ab?+b*. 
Ex. 3. Required the cube of 2a—1. 


INVOLUTION AND POWERS. 109 
Ld. Sda¥~joeah 4+ 

Ex. 4. Required the fourth power of 3a-h. 

Ex. 5. Required the square of atb+e. -% 0. 2°? +e 

Hence it appears that the square of a trinomial is common 
of the sum of the squares of all the terms, together with twice the 
sum of the products of all the terms multiplied together two and 
two. 

Ez. 6. Required the cube of 2Qab+cd. 


Ex. 7. Required the pent aka of eUn 
Ex. 8. Required the cube oF ats 


Ex. 9. Required the cube of n+. x54 


Ex. 10. Required the square of a+b+c+d+e. 

From this example we infer that the square of any polynomial 
is composed of the sum of the squares of all the terms, together 
with twice the sum of the products of all the terms multiplied to- 
gether two and two, and this proposition may be rigorously 
demonstrated. 

It is obvious that this rule for a polynomial includes the pre- 
ceding rule for a trinomial, and that in Art. 60 for a binomial. 


% 
aa © 


y ~ 
.~ 


SECTION XI. 





EVOLUTION AND RADICAL QUANTITIES. 


(143.) The square root of a quantity is a factor which, multi- 
plied by itself once, will produce that quantity. 

Thus, the square root of a’ is a, because a when multiplied 
by itself produces a’. 

The square root of 144 is 12 for the same reason. 

According to Art. 22, the square root is indicated by the 


sign V 
Thus, Va=a, 
and V 144a?=12a. 


(144.) According to Art. 137, in order to square a monomial, 
we must square its coefficient, and multiply the exponent of 
each of its letters by 2. Therefore, in order to derive the 
square root of a monomial from its square, we must 

I. Extract the square root of its coefficient. 

Il. Divide each of the exponents by 2. 

Thus we shall have 

Vv 64a'b'=8a'd’. 
This is manifestly the true result, for 
(8a°b’)?=8a'l’ x 8a'h’?=64a'D". 

So, also, 

| V 625a°b'c' =25ab'c’.. 

For, (25ab'c’)?=25ab'c* X 25ab'‘c’, 

=625a’b*c°. 

1. Required the square root of 196a*b‘c*d’. | 

2. Required the square root of 225a*"b""z". L 

(145.) According to Art. 140, a fraction is involved by i in- 
volving both the numerator and denominator ; hence it is ob- 


EVOLUTION AND RADICAL QUANTITIES. 111 


vious that the square root of a fraction ts equal to the root of 
the numerator divided by the root of the denominator. 


a’ 


Thus the square root of © 5 is - 


2 


1. Find the square root of rarer 


9a°b' 
16c°d’* 

(146.) It appears, from Art. 144, that a@ monomial can not 
be a perfect square unless its coefficient be a square number, and 
the exponents of its letters all even numbers. 

Thus, 7ab’ is not a perfect square, for 7 is not a square num- 
ber, and the exponent of a is not an even number. Its square 
root may be indicated by the usual sign, thus, V7ab?. Ex- 
pressions of this nature are called surds, or radicals ai the sec- 
ond degree. 

(147.) We have seen; Art. 137, that whatever may be the 
sign of a monomial, its square must be affected with the sign 
+. Hence we conclude that 

If a monomial be positive, its square root may be either post- 
tive or negative. 

Thus, / 9a'=+3a’, or —3a’, 
for either of these quantities, when multiplied by itself, pro- 
duces 9a*. We therefore always affect the square root of a 
quantity with the double sign +, which is read plus or minus. 


Thus, | Jaa =+2a' 
J 25a°b'=+5ab’. | 

(148.) If a monomial be affected with a negative sign, the 
extraction of its square root is impossible, since we have just 
seen that the square of every quantity, whether positive or 
negative, is necessarily positive. 

Thus, Vv —4, / —Y, V/ — 5a, 
are algebraic symbols representing operations which it is im- 
possible to execute. Quantities of this nature are called im- 
aginary or impossible quantities, and are symbols of absurdity 
which we frequently meet with in resolving quadratic equa- 
tions. 


2. Find the square root of ——, 


wir. 
% 


112 EVOLUTION AND RADICAL QUANTITIES. 


Such quantities may be represented by the form 


—_—_—. 


¥ —a, which equals 
Vax—1=Vav—l. 

So that VavV—1 is a general form for all imaginary quan- 
tities of the second degree. Thus, 

V—4 =vV4 X—1= 2 V-1, 
V—9 =V9 xXx—-1= 3 V-1, 
V —5a= V5aX —1= Vbav—1. 

That is, the square root of a negative quantity may always be 
represented by the square root of a positive quantity multiplied 
by the square root of —1. 

(149.) According to Art. 138, in order to raise a monomial 
to any power, we raise the numerical coefficient to the given 
power, and multiply the exponent of each of the letters by the 
exponent of the power required. Hence, reciprocally, to ex- 
tract any root of a monomial, we obtain the following 


RULE. 


I. Extract the root of the numerical coefficient. 

Il. Divide the exponent of each letter by the index of the re- 
quired root. 

Thus, V 64a°bD* =4a’°b. 

Y16b"c"=2'c*. 

From Art. 145, it is obvious that to extract any root of a 
fraction, we must divide the root of the numerator by the root 
of the denominator. 

27a°h®. 3a°b 


Thus the cube root of ——— is ——; 
nat 39% Bry Qa" 


which may be written Sabo 'y— 


(150.) Let us now consider the sign with which the root 
should be affected. We have seen, Art. 139, that every even 
power is positive, but an odd power has the same sign as its 
root. 

Thus — a, when raised to different powers in succession, 
will give | 
—da, +a’, —a’, +a‘, —a’, +a’, —a’, &e. ; 


hye 


EVOLUTION AND RADICAL QUANTITIES. 113 


and +a, in like manner, will give 
+a, +a’, +a’, +a°, +a°, +a°, +a’, &c. 

Since every even number may be expressed by 2, every even 
power may be considered as the square of the nth power, or 
a*"=(a")’, and must, therefore, be positive ; and, in like manner, 
since an odd number may be expressed by 2n+1, every power 
of an uneven degree may be considered as the product of the 
2nth power by the original quantity, and must, therefore, have 
the same sign with the monomial. 

Hence it appears, 

I. An odd root of any quantity must have the same sign as the 
quantity itself. 

Thus, «/ +8a°= +2a. 

V —8a°=— 2a. 

Y —32a"b'=—2a°b. 

57 +32a"b'= +2a°D. 
II. An even root of a-negative quantity 1s ambiguous. 
Thus, 81a‘b!=+38ab’. 


¥ 64a" =+2a'. 
Ill. An even root of a negative quantity is impossible. 
For no quantity can be found which, when raised to an even 
power, can give a negative result. 


Thus, —a, /—b, are symbols of operations which can 
not be performed, and they are therefore called impossible or 


imaginary quantities, as / —a, in Art. 148. 


EXAMPLES. 


1. Find the fourth root of 81a’. 
Ans, +3a’. 

2. Find the fifth root of —248a"b’c—™. - 
3. Find the cube root of neti fb 

a’ 
9x Se 
32a"b— 

243 

(151.) According to the rule of Art. 149, we perceive that, 
in order that a monomial may be a perfect power of any degree, 
its coefficient must be a perfect power of that degree, and the 


4. Find the square root af: 


5. Find-the fifth. root of ———— 


114 EVOLUTION ANP RADICAL QUANTITIES, 


exponent of each letter must be'divisible by the index of the 
root. | 

When the quantity whose root is required is not a perfect 
power of the given degree, we can only indicate the operation 
to be performed. Thus, if it be required to extract the cube 
root of 4a°b’, the operation may be indicated by writing the 
expression thus, 

VY 4a°b'. 

Expressions of this nature are called surds, or irrational 
quantities, or radicals of the second, third, or nth degree, ac- 
cording to the index of the root required. 

(152.) The method of extracting the roots of polynomials 
will be considered in Section XVII. There is, however, one 
class so simple and of so frequent occurrence that it may prop- 
erly be introduced here. In Arts. 60 and 61 we have seen that 
the square of a+b is a’+2ab+b’, 
and the square of a—b is a’—2ab+b’. 

Therefore, the square root of @+2ab+D’ is ab. 

Hence a trinomial is a perfect square when two of its terms 
are squares, and the third is the double product of the roots of 
these squares. 
~ Whenever, therefore, we meet with a quantity of this de- 
scription, we may know that its square root is a binomial; and 
the root may be found by extracting the roots of the two terms 
which are complete squares, and connecting them by the sign of 
the other term. 

Ex. 1. Find the square root of a’+4ab+ 4b’. 

The two terms, a’ and 4b° are complete squares, and the 
third term 4ab is twice the product of the roots a and 2b; 
hence a+2b is the root required. 

Ex. 2. Find the square root of 9a’—24ab+160’*, 

Ez. 3. Find the square root of 9a*—30a*°b+25a’b’, 

Ex. 4. Find the square root of 4a’+ 14ab+90’. 

(153.) No binomial can be a perfect square. For the square 
_ of a monomial is a monomial; and the square of a binomial 
consists of three distinct terms, which do not admit of being 
reduced with each other. 

Thus such an expression as 

a+b? 


IRRATIONAL QUANTITIES. 115 


is not a square; it wants the term +2ab to render it the square 
of a+b. This remark should be continually borne in mind, 
as beginners often put the square root of a’+0’ equal to a+b. 


IRRATIONAL QUANTITIES, OR SURDS. 


(154.) A rational quantity is one which can be expressed in 
finite terms, and without any radical sign; as a, 5a’, &c. 

Irrational quantities, or surds, are quantities affected with a 
radical sign, and which have no exact root, or a root which can 
be exactly expressed in numbers. 

Thus, 3 is a surd, because the square root of 3 can not be 
expressed in numbers with perfect exactness. 

In decimals it is 1.7320508 nearly. 

(155.) We have seen, Art. 144, that in order to extract the 
square root of a monomial, we must divide each of its expo- 
nents by 2. 

Thus the square root of @ is a or a; that of a‘ is a’; that 
of a’ is a’, and so on; and as this principle is general, the square 


3 5 
root of a must necessarily be a*, and that of a® must be a? ; 


1 
and, in the same manner, we shall have a’ for the square root 
of a’. Whence we see that 


1 
a* is equal to /a, 


)(%) 


a’ is the same as Va’, 


ls 


a? is equivalent to Va’, 
&c., &c. 


We have also seen, Art. 149, that in order to extract any 
root of a monomial, we must divide the exponent of each letter 
by the index of the required root. 

Thus, the cube root of a is a’, or a; the cube root of a’ is 
a’; the cube root of a® is a’, and so on. So, also, the cube 


2 HOlM4e 
root of a’ is a’; the cube root of a* is a*; the cube root of a, 
1 
or a’,isa®. Whence it appears that 
1 apa 
a® is the same as Va, 


i : a 
a’ is equivalent to Va’. 


iv 


116 IRRATIONAL QUANTITIES. 


ais equivalent to Va", 
&c., GEG. 


. 1 — ‘ i rs 
In the same manner, the fourth root of a is a*, which expres- 


sion has therefore the same value as Va; the fifth root of a4 


will be a’, which is, consequently, equivalent to Ya, and the 
same principle may be extended to all roots of a higher de- 
gree. | 

(156.) Other fractional exponents are to be understood in 


5 
the same way. Thus, if we have a‘, this means that we must 
first take the fifth power of a, and then extract its fourth root; 


ed fas: 
so that a* is the same as Va’. 


m 


So, also, to find the value of a", we must first take the mth 
power of a, which is a”, and then extract the nth root of that 


m 


power; so that a” is the same as Va”. 
Hencté the numerator of a fractional exponent denotes the 
power, and the denominator the root to be extracted. 


e . e 1 
Again, let it be required to extract the cube root of * 
4 


bag! 


] 
In the first place, ane. Now, to extract the cube root 


of a—*, we must divide its exponent by 3, which gives us 


ae. 


1 1 
But the cube root of Gi may also be represented by —. 
as 
] 


Ls 
a 


4 
Hence ~7 is equivalent to a °*. 


—_, 


So, also, 7 is equivalent toa ”, 


1 
is equivalent toa ", 


m 
is equivalent to a "*. 


8 al = §.-l- &, 


Thus we see that the principle of Art. 69, that a factor may 
be transferred from the numerator to the denominator of a 


IRRATIONAL QUANTITIES. 117 


fraction, or from the denominator to the numerator by chang- 
ing the sign of its exponent, is applicable also to fractional ex- 
ponents. . 

We may therefore entirely reject the radical signs hitherto 
made use of, and employ, in their stead, the fractional expo- 
nents which we have just explained ; and, indeed, many of the 
difficulties in the reduction of radical quantities disappear when 
fractional exponents are substituted for the radical signs. 


PROBLEM I. 

To reduce surds to their most simple forms. 

(157.) Surds may frequently be simplified by the application 
of the following principle: the square root of the product of two 
or more factors is equal to the product of the square roots of 
those factors. 

Or, in algebraic language, 

Vab= Vax vd. 

For each member of this equation squared will give the 
same quantity. 

Thus, the square of Vab is ab. 

And the square of Vax /b is (/a)’?X (/b)?=ab. 

Hence, since the squares of the quantities Vab and yaX Vb 
are equal, the quantities themselves must be equal. 


Let it be required to reduce Vv 4a to its most simple form. 


This expression may be put under the form /4X Va. 
_ But 4 is equal to 2. 


to] = 


Hence, V4a= /4X /a=2/a=2a’. 
2,/a is considered a simpler form than V 4a, for reasons which 
will be better understood hereafter. 

Again, reduce ,/48 to its most simple form. 

/48 is equal to V16X3= V16X /3=4 V3. 

Therefore, in order to simplify a monomial radical of the 
second degree, separate it into two factors, one of which is a 
perfect square ; extract its root; and prefix it to the other factor 


- with the radical sign between them. 


In the expressions 2,/a and 4,/3, the quantities 2 and 4 are 
ealled the coefficients of the radical. ‘ 


118 IRRATIONAL QUANTITIES. 


EXAMPLES. 
1. Reduce 2./32 to its most simple form. 

Ans. 8/2. 
2. Reduce V125a* to its most simple form. 


Ans. 5a Vv 5a. 
8. Reduce V98ab‘ to its most simple form. 


Ans. 7b’ V 2a. 
. Reduce V294ab’ to its most simple form. 
. Reduce 7-V80abe* to its most simple form. 
. Reduce V98a’x"y’ to its most simple form. 


IO oO 


. Reduce v45a’b*c'd to its most simple form. KES 

8. Reduce V864a7b'c" to its most simple form. 

(158.) Surds of any degree may be simplified by the appli- 
cation of the following principle, which is merely an extension 
of that already proved in the preceding Article. 

_ The nth root of the product of any number of factors is equal 
to the product of the nth roots of those factors. 

Or, in algebraic language, 

Vab=Vax Vb. 
For, raise each of these expressions to the nth power, and we 
shall obtain the same result. 

Thus, the ath power of Vab is ab. 

And the nth power of Vax Vb is (Va)"x (Vb)"=ab. 

Hence, since the same powers of the quantities Vab and 
Vax Vb are equal, the quantities themselves must be equal. 

Let it be required to reduce V8qa’ to its most simple form. 

This is equivalent to V8x Va’, which is equal to 2 Va’. 

Again, take the expression 

V 48a’. 

This is equivalent to /16a'x 4/3a, which is equal to 2a 3a. 

Hence, to simplify a monomial radical of any degree, we 
have the following 


RULE. 


* Separate the quantity into two factors, one of which is an ex- 


RRATIONAL QUANTITIES. ~ 119 


act power of the same name with the root; extra. .ts root; and 
prefix it to the other factor with the radical sign between them. | 
‘In the expressions 2a’ and 2a4/8a, the quantities 2 and 2a 
placed before the radical sign are called the coefficients of the 
radical. 


EXAMPLES. 


1. Reduce /56a°b’ to its most simple form. 
Ans. 2ab?V Ta’. 
2. Reduce V54a‘b‘c’ to its most simple form. “ne 
Ans. 8ab V 2ac’. 
3. Reduce V48ua'‘h*c’ to its most simple form. 
Ans. 2ab’c ¥ 8ac’. 
4. Reduce /192a"bc” to its most simple form. 
5. Reduce V192a‘b’c* to its most simple form. 
6. Reduce 9810? to its most simple form. 
(159.) There is another principle which can frequently be 
employed to advantage in simplifying radicals. 
The square of the cube of a is equal to the sixth power of a. 
For the square of the cube of a is a’*xXa’, 
which equals a*t°=a’, 
So, also, the fourth power of the cube of a is equal to the 
twelfth power of a. 
For (a)i=aXe Xa xa 


And, in general, the mth power of the nth power of any 
quantity is equal to the mnth power of that quantity. 
That is (a")"=a™. 
Hence, conversely, | 
The math root of any quantity is equal to the mth root of 
the nth root of that quantity. 
Thus, the fourth root = the square root of the square root ; 
“ the sixth root = the square root of the cube root, or 
the cube root of the square root; 
the eighth root = the square root of the fourth root, or 
the fourth root of the square root; 
“ the ninth root = the cube root of the cube root. 


6é 


120 IRRATIONAL QUANTITIES. 


Hence, when the index of a root is the product of two or more 
factors, we may obtain the root required by extracting in suc- 
cession the roots denoted by those factors. 

Ex. 1, Let it be required to extract the sixth root of 64. 

The sixth root is equal to the cube root of the square root. 

The square root of 64 is 8, 
and the cube root of 8 is 2. 

Hence the sixth root of 64 is 2. 

Ex. 2. Let it be required to extract the eighth root of 256. 

The eighth root is equal to the fourth root of the square root; 
or to the square root of the square root of the square root. 

The square root of 256 is 16, 
and the fourth root of 16 is 2. 

Hence the eighth root of 256 is 2. 

When one of the roots can be extracted, and the other can 
not, a radical may be simplified by extracting one of the roots. 

Thus, the fourth root of 9 is equal to the square root of the 
square root of 9; that is, the square root of 3. 

Or, algebraically, V9= V3. 

Ex. 3. Reduce 44a’ to its most simple form. 

Ans. ¥2a. 

Ex. 4. Reduce V36a’l’ to its most simple form. © ©“ 

Ex. 5. Reduce “a” to its most simple form. V 


Ez. 6. Reduce V25a'‘b’c’ to its most simple form. 


PROBLEM II. 


(160.) To reduce a rational quantity to the form of a surd. 
The square root of the square of a is obviously a; that is, 
nn 
; T= Vt Oa 
So, also, the cube root of the cube of a is a; 
3 
that is, i=Vi=ai. 
Hence, to reduce a rational quantity to the form of a surd, 
we have the following 


RULE. 


Raise the quantity to a power of the same name with the given 
root, and then apply the corresponding radical sign. 


cs 


IRRATIONAL QUANTITIES. 12] 


EXAMPLES. 


1. Reduce 3 to the form of the square root. 
Here 3X3=3?=9; whence 3= /9. Ans. 
2. Reduce az to the form of the square root. 

| Ans. Va'x’, or (a°x")? 

3. Reduce 2x" to the form of the cube root. 
Ans. ¥/82". 
Reduce 5+0 to the we of the square root. 
. Reduce —38z to the form of the cube root. 
. Reduce —12” to the form of the fourth root. 
Reduce a‘b* to the form of the square root. 
. Reduce a” to the form of the nth root. . 

It will be observed, that this Problem is nearly the reverse of 
the preceding, and, consequently, brings quantities into a less 
simple form ; nevertheless, this form is sometimes better suited 
to subsequent operations, as will be seen hereafter. 


2rx are 


PROBLEM III. 


(161.) To reduce surds which have different indices to others 
of the same value dais a common index. 


Ez. 1. Reduce a? and a® to surds having the same radical 
sign. 

From the preceding Article, it is obvious that the square 
root of a is equal to the sixth root of the cube of a; 


3 aaceal 

that is, aa hah 
1 2 

So, also, aro a* 

1 

Thus, the quantities a” 

which are of the same value, and have the common index 6. 


1 t 
Ez. 2. Reduce 3? and 2° to a common index. 
ak 3 ery it 1 
HOE ed 6shp tere ie 
2328 — (2*)8 4°, 


Hence 27 and V4 are the quantities required. 
Whence we derive the following 


Liss IRRATIONAL QUANTITIES. 


ba 
RULE. 


Reduce the fractional exponents to a common denominator ; 
raise each quantity to the power denoted by the numerator of its 
reduced exponent; and take the root denoted by the common de- 
nominator. 


1 1 
Ex. 8. Reduce 2? and 4* to a common index. _ 
Ans. V4 and V8. 
1 ay } 
Ex. 4. Reduce a’? and a? to a common index. 


1 2 
Ex. 5. Reduce a? and b? to a common. index. 


2 3 
Ex. 6. Reduce 5° and 7* to a common index. 
1 1 


Ex. 7. Reduce a” and b” to a common index. 


PROBLEM IV. 


To add surd quantities together. 

(162.) Two radicals are similar when they have the same 
index, and the same quantity under the radical sign. 

Thus, 3Va and 5Va are similar radicals. 

So, also, 7Vb and 10 V8 are similar radicals. 

But Va and Va are not similar radicals; for, although they 
have the same quantity under the radical sign, they have not 
the same index. 

Ez. 1. Find the sum of 2Va and 3Va. 


As these are similar radicals, we may unite their coefficients 
by the usual rule ; for it is evident that twice the square root of 
a and three times the square root of a make five times the 
square root of a. Hence the following 


RULE. 


When the radicals are similar, add the coefficients, and annex 
the radical part. | 
But tf the quantities are dissimilar, and can not be made sim- 
ilar by the reductions in the preceding Articles, they can only be 
connected together by the sign of addition. 
Ex. 2. Add V6 to 2V6. 
Ans. 3V6. 


IRRATIONAL QUANTITIES. 123 


Ex. 3. Add 5Va and —2¥a. 

Ex. 4. Add aVb+e and «Vb-+e. | 

If the radical parts are originally different, they must, if pos- 
sible, be made alike by the preceding methods. 

Hix. 5. Add ¥27 to Vv 48. 


Here V27=V 9X3=3 V3, 
and V48= V16X3=4 V3. 
Whence their sum =I V3. 
Ex. 6. Add together V500 and V108. 
Ans. 84/4. 
Ex. 7. Add together 4V147 and 3V75. 
Ans. 48 V3. 
Ex. 8. Add together 3 V2 and 2vV 74. 
Here 3/2 =38/it =3 10, 
and 2/ J H=2V dH Fs V10. 
Whence their sum = 4/10. 


Ez. 9. Add together /72 and v128. 

kz. 10. Add together Y180 and 405. 
fiz. 11. Add together V40 and VY 135. 
Ex. 12. Add together 8 732 and 5¥2. 


PROBLEM V. 


To find the difference of surd quantities. 

(1638.) It is evident that the subtraction of surd quantities 
may be performed in the same manner as addition, except that 
the signs in the subtrahend are to be changed according to 
Art. 43. | 

Ex. 1. Required to find the difference between 448 and 
V112. 


Here /448= V64X7=8,/7, 

and J112= V16X7=4,/7. 

Whence the difference i ae 
Kx. 2. Find the difference between 192 and 7/24. 
Here V192= 764x3=493, 

and V¥24 =V 8x3=2¥3. 





Whence the difference =273. 


124 IRRATIONAL QUANTITIES. 


Ex. 3. Find the difference between 5/20 and 3/45. 


Here 5V20=5V4xX5=10V5, 
and 3745= 3V9X5=_ 9V5. 
Whence the difference — V5, 


Ex. 4. Find the difference between 2./50 and ./18. 

Ez. 5. Find the difference between 2% 320 and 3/40. 
Ex. 6. Find the difference between V80a‘z and Vv 20a’z*. 
Ex. 7. Find the difference between 2 V72a? and vV162a’. 


PROBLEM VI. 


Lo multiply surd quantities together. 

(164.) Let it be required to multiply Va by Vb. 

The product will be Vab. 

For if we raise each of these quantities to the power of n, 
we obtain the same result, ab; hence these two expressions 
are equal. We therefore have the following 


RULE. 


When the surds have the same index, multiply the quantities 
under the sign by each other, and prefix the common radical 
sign. If there are coefficients, these must be multiplied separ- 
ately. 

Ex. 1. Required the product of 3V8 and 2V6. 


Here 3V 8, 
multiplied by 2. Gis ae 
gives : 6 V48=6 V16X3=24V3. Ans. 


Proof. Square 3,/8, and we obtain 9X 8=72. 
Square 2./6, and we obtain 4x 6=24. 
72 multiplied by 24=1728. 
Also, 24./3 squared =576 X 3=1728. 
Ez. 2. Required the product of 5V8 and 3V5. 
Ans. 30V 10. 
Ex. 3. Required the product of 7718 and 54/4. 
Ans. 70/9. 
Ex. 4. Required the product of 1/6 and 2,Y17. 7% 
Ex. 5. Required the product of 1/18 and 5720. J°« ys 


Savr 


IRRATIONAL QUANTITIES. | 125 


In the preceding examples, let all the results be reduced to 
their simplest form. 

If the surds have not the same index, they must first be re- 
duced to a common index, by Art. 161. 


Ex. 6. Required the product of v2 and Y3. 


Here | 2=2=Q2)'=V 8, 
and ¥3=3'=(3)"=V 9. 
Whence the product = Vv 72. 


(165.) We have seen, in Art. 50, that powers of the same 
quantity may be multiplied: by adding their exponents. The 
same principle may be extended to roots of the same quantity. 


Let it be required to multiply Va by Va, or a? by a’. 


1 
2 


3 oF 2 
We have seen, in Art. 161, that a2=a®, and a?=a’. 
3 1 L 1 
Bura'=-a° <a" xa", 
3 1 1 
and * a°=a" xa’: 
‘ 1 1 1 1 1 5 
The product, therefore, is a® xa*xa° xa® xa°=a’. 
Hence, roots of the same quantity may be multiplied by add- 
ing their fractional exponents. 
1 1 
Ez. 1. Multiply 5a? by 3a’. 
5 
Ans. 15a°. 
2 1 | 
Ez. 2. Multiply 3a* by 21a’. 
ie Is 
Ez. 3. Multiply 32*y* by 4z?y°. 
1 z 
Ex. 4. Multiply (a+0)" by (a+b)". 
(166.) If the rational quantities, instead of being coefficients 
of the radical quantities, are connected with them by the signs 


+ or —, each term of the multiplier must be multiplied into 
each term of the multiplicand. 


1. Let it be required to multiply 8+ v5 


by 2— V5 
6+2/5 

—3./5—5. 

We obtain the product 6— /5—5, 


which reduces to 1— 5. 


126 IRRATIONAL QUANTITIES. 


2, Multiply 742.6 by 9-5 v6. 
Ans. 3—17,/6. 
3. Multiply 9+-2./10 by 9-210. 
Ans. 41. 


PROBLEM VII. 


To divide one surd quantity by another. 

(167.) Let it be required to divide Ya’ by Ya’ 

The quotient must be a quantity which, multiplied by the 
divisor, shall produce the dividend; we thus obtain Va; for, 
according to Art. 164, Va’ x 'Va= Ja! ; 


vo. 
that is, ee = Va, 
Hence the following 
RULE. 


Quantities under the same radical sign may be divided like 
rational quantities, the quotient being affected with the common 
radical sign. If there are coefficients, they must be divided 
separately. 


If the radicals have not the same index, we must first vetaaiee 


them to a common index. 





EXAMPLES. 
1. It is required to divide 8/108 by 2/6. 
8/108 ae 
Here 276 =4,/18=249/ 9212/2, Ans. 
2. Divide 8 V512 by 4/2. 
84/512 EYE r= a | 
Here ae = V256=2V64xX4=8V4. Ans. 
3. Divide 6/54 by 3 V2. 
Ans. 6. 

4, Divide 4V72 by 24/18. 
5. Divide 4V6a’y by 2V3y. 

Ans. 2av2. 


6. Divide 16(a°b)™ by 8(ac)”. LY ais 
7. Divide 4/12 by 23. IVs 


IRRATIONAL QUANTITIES. 127 
As the radicals in this last example have not the same in- 
dex, they must be reduced to.a common index. 
1 2 1 
4 12=4(12)*=4(12)°=4(144)*. 
‘e 3 1 
2v8 =2(8)' =2(8)' =2(27)". 
4(144)" 
2(27)° 
(168.) We have seen, in Art. 67, that, in ete to divide 
quantities expressed by the same letter, we must subtract the 


3 exponent of the divisor from the exponent of the dividend. 
The same principle may be extended to fractional exponents. 


“4 





Hence =2(144)8— 2(1) SFO V8, 


4 z 
Thus, let it be required to divide a? by a’. 
According to the <dtvta ae 


4 
a* ‘at 
pars — 6 <i 
a? 


Hence a root is divided se i another root of the same letter or 
quantity, by subtracting the exponent of the divisor from that 
of the dividend. 

Ex. 1. Divide (ab)* by (ab)*. 

Ans. (ab)>. 

Er. 2. Divide a* by a you 

Ez. 3. Divide a* ® by a : : 


. Ex. 4. Divide a” by a, 
e- Ex. 5. Divide 4,/ab by 2Vab. 
2. Ans. 2/ ab. 
Nee: 


PROBLEM VIII. 


(169.) To raise surd quantities to any power. 


Let it be required to find the square of a’. 
The square of a quantity is found by multiplying it by itself 
once. 


1 tat Hu 

Hence the square of a® is equal to a?Xa®=a*’*=a'. 
\y2 2 
That is, (a?) he 


128 IRRATIONAL QUANTITIES. 


Again, let it be required to find the cube of a®. 
The cube of a quantity is found by multiplying it by itself 
twice. 


1 1 1 1 3 
Hence the cube of a® is equal to a®xa*xXa°=a’ ; 


that is, (a3) ; aa 
1 n 


In the same manner we should find the nth power of a"=a”. 
Hence we have the following 


RULE. 


Radical quantities are involved by multiplying their fractional 
exponents by the exponent of the required power. 


Ex. 1. Required the fourth power of 243, 
Ex. 2. Required the cube of 2 v3. 
Ans. £3. 
Ex. 8. Required the square of 373. 7 
Ex. 4. Required the cube of 17 V21. Vad 
Ez. 5. Required the fourth power of 1/6. 
| Ans. 5. 
(170.) If the radical quantities are connected with others by 
the signs + and —, they must be involved by a multiplication of 
the several terms. 
fiz. 1. Required the square of 3+ 5. 
3+ Vd 
9435 
3/5+5 
The square is - 9+6V/5+5 
or 1446/5. Ans. 
Ex. 2. Required the square of 3+2,/5. LGHSA 
These two examples. are comprehended under the rule in 
Art. 60, that the square of the sum of two quantities is equal 
to the square of the first, plus twice the product of the first by 
the second, plus the square of the second. | 
Ex. 8. Required the cube of yx+38 yy. x AE VAN FG 5 
Ex. 4. Required the fourth power of y3— 2. 
Ans. 49—20 V6. 





IRRATIONAL QUANTITIES. 129 


PROBLEM IX. 
To find the roots of surd quantities. 


(171.) A root of a quantity is a factor which, multiplied by 
itself a certain number of times, will produce the given quan- 
tity. But we have seen that a radical quantity is involved by 
multiplying its exponent by the exponent of the required pow- 
er. Hence, 

To find the roots of surd quantities, 


Divide the fractional exponent by the index of the required 
root. 


Thus, the square root of a? is at gt. 
For, by Art. 169, we obtain the square of a by multiplying 
the exponent 3 by 2; 
that is, (a?) <0 <q", 
EXAMPLES, 
1. Find the square root of 9(3)°. 
lis 


Here (9(3)3)?=9? x 3° —3(3)"=3 V3." Ans. 

2. Required the cube root of 1/2. 
Ans. 1/2. 

. Required the square root of 10°. cs 

. Required the cube root of a‘. 


2 
. Required the fourth root of 1£a’*. 
. Required the cube root of {44,a°. 


tO oO FP & 


. Required the cube root of $4/}. if 
Ans. V3- 


PROBLEM X. 
To find multipliers which shall cause surds to become rational. 
(172.) I. When the surd is a monomial. 
The quantity ya is rendered rational by multiplying it by 
Jf a. 


a 
For J/ax Ja=a'x a=. 
I 


130 IRRATIONAL QUANTITIES. 


1 bee Wat i 3 
So, also, a* is rendered rational by multiplying it by a’. 
Livbe B15 98 
For a X<aleHa' =a. ey 
1 3 
Also, a‘ is,rendered rational by multiplying it by a*. 


i 3 4 
For @ x0. — asa, 
n-1 
In genni a” is rendered rational by multiplying it by a” 
SMD gray cir tdh 2 
For OX ibis Chih ese. 


Hence we deduce the following 


RULE. 


Multiply the surd by the same quantity having such an ex- 
ponent as, when added to the exponent of the given surd, shall be 
equal to unity. 

(173.) Il. When the surd is a binomial. 

If the binomial contains only the square root, multiply the 
given binomial by the same expression with the sign of one of its 
terms changed, and it will give a rational product. 

Hix. 1. The expression Va+ Vb 


Multiplied by Ja— Jb 

a+ Vab 

—Vab—b 

Gives a product a —b, which is rational. 
Ez. 2. Find a multiplier which shall render 5+ ./3 rational. 
Given surd, 5+ /38 
Multiplier, 5— /3 
Product, 25—3=22, as required, 


These two examples are comprehended under the Rule in 
Art, 62, the product of the sum and difference of two quarries 


is equal to the difference of their squares. 


Ez. 3. Find a multiplier that shall Hue /5+ 4/3 rational, 
and determine the product. , ey ae 


Ex. 4. Find a multiplier that shall make yv5— y2 rational, 


and determine the product. eh a emg 


Ex. 5. Find a multiplier that dee make Va— Vabe ra- 
tional. vs PA 


IRRATIONAL QUANTITIES. 131 


Ill. When the surd is a trinomial, it may be reduced, by 
successive multiplications, first to a binomial surd, and then to 
a rational quantity. 

Ez. 1. Find multipliers that shall make y5+ /38— 2 ra- 
tional. 








Given surd, J/d+ /38— V2 
First multiplier, yv5+ /3+ /2 
5+ /15—/10 
vio Si74/6 

LO sO 
First product, 2/15+ 6 
Second multiplier, 2/15— 6 

60412715 

—12./15—36 

Second product, 60—36=24, a rational quantity. 


Hx. 2. Find multipliers that shall make va+ yVb4+ Ve ra- 
tional, and determine the product. 


— 
yf 


, 1) I eA 
wo Ot dh lt 


be i 
pa 


PROBLEM XI. 
(174.) To reduce a fraction containing surds to another hav- 
ing a rational numerator or denominator. 
RULE. 


Multiply both numerator and denominator by a factor which 
will render either of them rational, as the case may require. 


Ex. 1. If both terms of the fraction a be multiplied by 
/a, it will become cmt in which the numerator is rational. 


/ab 


V ab 
Or if both terms be multiplied by 0, it will become - 





, 10 
which the denominator is rational. 

Ex. 9. Reduce the fraction 3 to one that shall have a ra- 
tional denominator. 


2/3 
Ans. Saas 


132 IRRATIONAL QUANTITIES, 
4 el 








: ‘ | ] ° F ie = . 
He 3. Reduce VEuwgs to a fraction having i rational de- 
nominator. * 
Beye 75 2 
eves v4 
o 
/2 ‘ : 2 
Ex. 4. Reduce 3 ye to a fraction having a rational de- 
nominator. 
3./2+2 
Ans. dalek des 
ae 
Ex. 5. Reduce a Sie to a fraction havi ti al d 
zt. 5. racti vin ° ~ 
! atv o7ing Ag Rapa alae 
nominator. 
Ez. 6. Reduce S to | havi 
x. 6. uce ———_—_ an ex 
Bt Vat pression having a 


rational denominator. 
Ans. 2+ ./2—/6. 


Ex. 7. Reduce /5+/2 to a fraction having a rational 
numerator. 


(175.) The utility of the preceding transformations may be 
illustrated by computing the numerical value of a fractional 
surd. | 

Ex. 1. Suppose it is required to find the square root of 3; 


that is, it is required to find the value of the fraction 


If we make the denominator rational, we shall have 





J2i1 . 
7 > in 
which it is only necessary to extract the square root of the 
numerator, and the value of the fraction is found to be 0.6546. 
_1¥5 
Vll+y3 


7/55—T/15 


Ex. 2. Itis required to find the value of the fraction 


Making the denominator rational, we have 


value of which is 3.1003. 


Ex, 3. Required the value of the expression FI4v8" 


Ans. 0.5595. 


IRRATIONAL QUANTITIES. 133 


Ez. 4. Required the value of the expression 
J/3 
2/843 V5—-TV2 


Ans. 0.7025. 
a . 9+2/10 | 
Ex. 5. Required the value of the expression 9-210 
Ans. 5.7278. 
PROBLEM XIL. MN be Fg 


(176.) To free an equation from radical quantities. | gt 
This may generally be done by successive involutions. For ~* 
this purpose, we first free the equation from fractions. If there 
is but one radical expression, we bring that to stand alone on 
one side of the equation, and involve the whole equation to a 
power denoted by the index of the radical. 
Ez. 1. Free the equation 
CHV ade) ; 
os a Rp 
from radical quantities, 
Clearing of fractions, and transposing a, we obtain 
° V 2ax+2°=ab—a, 
The square of this equation is 
2an+2°=a°b?—2a’b +a’, 
which is free from radical quantities. 
Ez. 2. Free the equation 
2a* 


nN a ere 


from radical quantities. 

If the equation contains two radical expressions, combined 
with other terms which are rational, it will generally be best 
to bring one of the radicals to stand alone on one side of the 
equation before involution. One of the radicals will thus be 
made to disappear, and, by repeating the operation, the re- 
maining radical may be exterminated. 

Ex. 3. Free the equation 

Vata+ Vb+y=c 
from radical quantities. 


134 IRRATIONAL QUANTITIES, 


Transposing one of the radicals, we obtain ; : 
Vatx=c— Vb+y. eo 
Squaring, we have 
ata=c'—2cVb+y+b+y. 
Transposing, so as to bring the radical to stand alone, we 
have 
2c Vb+y=c+b+y—a—z, 
which may be freed from radicals by squaring a second time. 
Sometimes the two radicals may be of such a form that it 
is best to bring both to the same member of the Bon be- 
fore involution. 
When an equation contains several radical quantities, it may 
generally be freed from them by successive involutions, but 
the best mode of procedure can only be determined by trial. 
Ez. 4. Free the equation 
V2e+7+ V3r—18= Vic+1 
from radical quantities. ; 
Ans. 6x*—154—126=27+122+36. \ X= 9 
When an equation contains a fraction involving radical 
quantities in both numerator and denominator, it is sometimes 
best to render the denominator rational by. Problem XI. 
Ez. 5. Free the equation 
Vz+Var—a C—O ae, 
Va—Va—a t74 
from radical quantities. 
Multiply both terms of the first fraction by Vz+ V2z—a, and 
we have 





(Vir vr — lems 


z—(e«—a) «—a 





or (Vi+ Vz—ay=——, 


Extracting the square root, we obtain 
an 
Vz2+VvVz2—a=_——: 
Vxr—a 
Clearing of fractions, we have 
Va?—ax+x2—a=an, 
which is easily freed from radicals. 


—. 


ae oe 


IRRATIONAL QUANTITIES. 135 


Ex. 6. Free the equation 
E+J/z xu’—z 
z—/xr 4 
frp radical quantities. 





Ans. v’>-4r+4=2. 
Ex. 1. Free the equation 


a Vz+l 5 


2+Ve+1 11 
from radical quantities. 





Ans. 9x°=647+64. 

Ez. 8. Free the equation 

Ve—zi—VbEZ -m 
V@e—-x#4+Vb+27 
from radical quantities. 

(177.) The preceding rules aa the reduction of radicals, are 
exact so long as we treat of absolute numbers, but require 
some modifications when we consider imaginary expressions, 
such as V—3, V—a, &c. 

Let it be required, for example, to determine the product of 
v—aby V—a. 

By the rule given in Art. 164, 

Vv —ax V—a= V—ax—a 
=v +a’. 

Now, V +a’=a, so that there is apparently a doubt as to 
the sign with which a ought to be affected in order to answer 
the question. However, the true result is —a, because any 
quantity must be equal to the square of its square root. 

That is,\v—ax V—a is the same as (V—a)’, and, conse- 
quently, is equal to —a. 

Again, let it be required to determine the product of ¥V—a 
by V—b. : 

By the rule in Art. 164. 

v—a —ax V—b=V—ax—b 
=V-+ab 
in a / 1, 

The result, however, is not properly ambiguous, and should 

be —Vab; for we have, according to Art. 148, 








136 IRRATIONAL QUANTITIES. 


V—a=Vva.v—l, 
and V—b=Vb.V—1. 
Hence Bre paella i,” 
V—ax V—b= Vab(V—1)? 
; =vVabx—1 
=~— Vvab. 


In the same manner we shall find for the different powers 
of V—1 the following results. 
¥—1 =v-—1, the first power. 
(Vv —1)?=—1, the second power. 
(V—1)'=—-1x V—I 
—— /—1, the third power. 
(v—1)'=(V—1)*x(V—1) 
=—-1x-1 
=+1, the fourth power. 

Since the four following powers will be found by multiply- 
ing +1 by the first, the second, the third, and the fourth 
powers, we shall again find for the four next powers 

+V—1, —1,-V-—1, +1; 
so that all the powers of V—1 will form a repeating cycle of 
these four terms. 

Whenever the student is at a loss to determine the product 
of two imaginary quantities, it is best to resolve each of them 
into two factors, one the square root of a positive quantity, and 
the other V—1, Art. 148. 


EXAMPLES. 
1, Let it be required to multiply v—§ ; by Vv—4. 
~ “Here we have vV—9=3V—1, 
and V—4=2V—1. 
Therefore, V—9X V—4=38V—1X2V—1 
=6 V (—1)' 
a= — 6, 


2. Multiply 1+ V—1 by 1-V—1. 

ee. Ans. 2. 
3. Multiply V18 by V —2. 
4. Multiply 5+2V—3 by 2—V—3. 


{ Pr Mew - 
~-_— 


~~ 


i™ 


“ 


SECTION XII. 


EQUATIONS OF THE SECOND DEGREE. 


(178.) According to Art. 96, quadratic equations, or equa- 
tions of the second degree, are those in which the highest power 
of the unknown quantity is a square. 

Quadratie equations are divided into two classes. 

I. Equations which involve only the square of the unknown 
quantity and known terms. These are called pure quadratics. 
Of this description are the equations 

ax’=b; 327°+12=150—2’, &c. 

They are sometimes called quadratic equations of two terms, 
because, by transposition and reduction, they can always be 
exhibited under the general form 

iar Or 

I. Equations which involve both the square and the first 
power of the unknown quantity, together with a known term. 
These are called affected or complete quadratics. Of this de- 
scription are the equations 

De i) 1S 


ax’+ba=c; 27—107=7; Bg tgs. 


They are sometimes called quadratic equations of three 
terms, because, by transposition and reduction, they can al- 
ways be exhibited under the general form 


ax’ +bz=c. 


PURE QUADRATIC EQUATIONS. 


(179.) The equation 
ax? =b 


~ 


138 EQUATIONS OF THE SECOND DEGREE. 


is easily solved. Dividing each member by a, it becomes 
Peis, 5 


z’=-. 


‘ a 
b 
Whence x =+,/ 


b ; 

If be a particular number, either integral or fractional, we 
can extract its square root either exactly or approximately by 
the rules of arithmetic. 

It is to be remarked, that since the square both of +-m and 


rages: b 
—m is +m’, so, in like manner, the square of Ay and that 


b b 
of si are both nes Hence the above equation is suscep- 


tible of two solutions, or has two roots; that is, there are two 
quantities which, when substituted for z in the original equation, 
will render the two members identical. These are 


b 
Tajo and Lay 
a as 


For, substituting each of these values in the original equation 
ax’=b, it becomes 


bye b : 
ax(+4/~) = De or ax-=b; 1:04 0D, 


b 2 
and ~ ax(— \/2) =b, or ax2=b 1. e., b=b. 


EXAMPLE I. 


Find the values of xz which satisfy the equation 
4¢°—7T=32'-+9. 

Transposing terms, 427°—3z°=9-+-7. 

Reducing, x’=16. 

Extracting the square root, 
Gaon/ L6eeck4, 

Hence the two values of z are +4 and —4, and they may ~° 

both be verified by substitution in the original equation. 
Thus, taking the first value, we have 


z 


EQUATIONS OF THE SECOND DEGREE. 139 


4X (+4)*"—7=3x (+4)?+9, 
or 4X 16—7=3X16+4+9; 
that is, 57=57. | 
Taking the second value of z, we have | 
4x (—4)'—7=8x (—4)"+9, 
_or 4x 16—7=3xX16-+9, as before. 


From the preceding examples we deduce the following 


RULE. 


Reduce the equation to the form ax*=b; then divide by the 
coefficient of x*, and extract the square root of both members of 
the equation. 

Ex. 2. Given z?—17=130—22’, to find the values of z. 





By transposition, an7= 147; 
therefore, z= 49, 
and Behan swig 
Ez. 3. Given z?+ab=5z’, to find the values of z. 
By transposition, ab=4z' ; 
therefore, + Vub=2e, 
+ Vab 
and c= ; 
2 
sole. a 
Ez. 4. Given z+ Va’?+2’?= +, to find the values of z. 
Va'+e 
Clearing of fractions, we obtain 2 Va’?+2°+a?+2°=2a’. 
By transposition, zVa’+z*=a’—2’. 
Squaring both sides, a’°2’+2'=a*—2a’a’* +2" ; 
therefore, fh ava=a; 
and So a": 
Op ke 
whence =; 
3 
a 
therefore, r= +t—., 
J/3 


2 
f 





5 
a =45, to find the values of z. 


Ex. 6. Given axv’—5c=bz’?—3c-+d, to find the values of z. 


; 42 
Ez. 5. Given 


« 


140 EQUATIONS OF THE SECOND DEGREE. 


nih OB 
Lez. 7. Given = 7m -3+57"= oat oan to find the values 
of z. L¢. tm? 
} v+3a2—7 
_ ix. 8. Given 2424187) to find the values of z. 
x 


Ans. 23} 


Clearing of fractions and transposing, we find in each mem- 
ber of this equation a binomial factor, which being canceled, 
the equation is easily solved. 


PROBLEMS. 


Prob. 1. What two numbers are those whose sum is to the 
greater as 10 to 7; and whose sum, multiplied by the less, pro- 
duces 270? 


Let 10z= their sum. 
Then 7x= the greater number, 
and 3x= the less. 
Whence 30z*=270, 
and r= 
therefore, cats, 


and the numbers are +21 and +9. 


Prob. 2. What two numbers are those whose sum is to the 
greater as m to n; and whose sum, multiplied by the less, is is 


Bhial toa? 
m(m—n) m 


Prob. 8. What number is that, the third part of whose 
square being subtracted from 20, leaves a remainder equal 
to 8? : . » ae 


Prob. 4. What number is that, the mth part of whose square 
being subtracted from a, leaves a remainder equal to b? 


Ans. + vVm(a—6). 
: ; ‘ Iv2 3 
Prob. 5. Find three numbers in the ratio of 5 3° and mt the 


sum of whose squares is 724. Abend ALT A 


- 


EQUATIONS OF THE SECOND DEGREE. 141 


Prob. 6. Find three numbers in the ratio of m, n, and p, the 


sum of whose squares is equal to a. 
Ans. 


Fe vets / ang ty/. aa 
m’+n*+-p m+n? +p" m+ n* +p" 
Prob. 7. Divide the number 49 into two such parts, that the 
quotient of the greater divided by the less, may be to the quo- 


tient of the less divided by the greater, as 4 to 3, 
Ans. 21 and 28. 


Prob. 8. Divide the number a into two such parts, that the 
quotient of the greater divided by the less, may be to the quo- 
tient of the less divided by the greater, as m to n. 

FP anes Sele dchios pala cetnadedd 
J/m+ /n J/m+/n 

Prob. 9. There are two square grass plats, a side of one of 
which is 10 yards longer than a side of the other, and their 
areas are as 25 to 9. What are the lengths of the sides? 


Prob. 10. There are two squares whose areas are as m to n, 
and a side of one exceeds a side of the other by a. What are 
the lengths of the sides ? 





a/m aJ/n 

F —— and, -—-__——, 

J/m— /n J/m—J/n 

Prob. 11. Two travelers, A and B, set out to meet each other, 
A leaving Hartford at the. same time that B left New York. 
On meeting, it appeared that A had traveled 18 miles more 
than B, and that A could have gone B’s journey in 152 hours, 
but B would have been 28 hours in performing A’s journey. 
What was the distance between Hartford and New York? 
Ans. 126 miles. 


Prob. 12. From two places at an unknown distance, two 
bodies, A and B, move toward each other, A going a miles more 
than B. A would have described B’s distance in 7 hours, and 
B would have described A’s distance in m hours. What was 
the distance of the two places from each other ? 


Ans 


J/m+/n 
J/m— f/n 
Prob. 13. A vintner draws.a certain quantity of wine out of 


Ans. ax 


‘ 


a. 


142 — EQUATIONS OF THE SECOND DEGREE. 
} 


a full vessel that holds 256 gallons; and then, filling the vessel 
with water, draws off the same quantity of liquor as before, 
and so on for four draughts, when there were only 81 gallons 
of pure wine left. How much wine did he draw each time? 
Ans. 64, 48, 36, and 27 gallons. 


Prob. 14. A number a is diminished by the nth part of it- 
self, this remainder is diminished by the nth part of itself, and 
so on to the fourth remainder, which is equal to b, Required 
the value of x. : 

Va 
Ans. Van¥Vb" 

Prob. 15. Two workmen, A and B, were engaged to work 
for a certain number of days at different rates. At the end of 
the time, A, who had played 4 of those days, received 75 shil- 
lings, but B, who had played 7 of those days, received only 
48 shillings. Now had B only played 4 days, and A played 7 
days, they would have received the same sum. For how 
many days were they engaged ? 

Ans. 19 days. 


Prob. 16. A person employed two laborers, allowing them 
different wages. At-the end of a certain number of days, the 
first, who had played a days, received m shillings, and the 
second, who had played b days, received n shillings. Now if 
the second had played a days, and the other b days, they 
would both have received the same sum. I*or how many days 
were they engaged ? 





COMPLETE QUADRATIC EQUATIONS. 
(180.) Suppose we have the equation 
z’—6*+9=1, 
in which it is required to find the value of z. 
Since each member of the equation is a complete square, if 


we extract the square root, we shall obtain a new equation — 


involving only the first power of x, which may be easily 
solved. 


EQUATIONS OF THE SECOND DEGREE. 1438 


We thus have x—3=+1, 

and, by transposition, 
t=3-l—4jior 2: 

In order to verify these values, substitute each of them in 
place of z in the given equation. ‘Taking the first value, we 
shall have 

4°—-§X4+9=1; 


that is, 16—24+9=1, an identical equation. 
Taking the second value of 2, we obtain 
2—6X2+9=1% 
that is, 4—12+9=1, an identical equation. 


Hence we see that a complete quadratic equation is readily 
solved, provided each member of the equation is a perfect 
square. But equations seldom occur under this form. Take, 
for example, 

z’—b2=—8. 

The preceding method seems to be inapplicable, because the 
first member is not a complete square. We may, however, 
render it a complete square by the addition of 9, which must 
also be added to the second member to preserve the equality. 
The equation thus becomes 

2? —62+9=9—8=1, 
which is the equation first proposed. 

The peculiar difficulty, then, in resolving complete equa- 
tions of the second degree, consists in rendering the first mem- 
ber an exact square. 

(181.) In order to discover a general method of solution, let 
us take the equation 

ax’ +br=c, 
which is the general form of equations of the second degree. 
We begin by dividing both members by a, the coefficient of x’. 
The equation then becomes 


: b C 
For convenience, let us put p= and Gas we shall then 
¢ 


have 
x + pr=q. 


Me 
tis 
: eG, - ey “ 
i asi . ‘ 
lig}? EQUATIONS OF THE SECOND DEGREE. 


We have seen that if we can by any transformation render 
the first member of this equation the perfect square of a bino- 
mial, we can reduce the equation to one of the first degree by 
extracting the square root. 

Now we know that the square of a binomial, +a, or 27+ 
2axr+a’*, is composed of the square of the first term, plus twice 
the product of the first term by the second, plus the square of 
the second term. 

Hence, considering z*+pz as the first two terms of the 
square of a binomial, and, consequently, px as being twice the 
product of the first term of the binomial by the second, it is — 


evident that the second term of this binomial must be a for 


2x EXa=pe. 
In order, therefore, that the expression z°*+-pz may be ren- 
dered a perfect square, we must add to it the square of this 


second term f ; that is, the square of half the coefficient of the 


first power of x; it thus becomes 


2 


x +po+4, 


P 
| 2° 
the left-hand member of the equation, in order that the equality 
may not be destroyed, we must add the same quantity to the 
right-hand member also; the equation thus transformed will 
become “ 

ae 
4 


Extracting its square root, we have 


2 
which is the square of «+ But since we have added f to 


2° pact 


Whence = P+ J qth 


We prefix the double sign +, because the square both of 


a 2 See 


EQUATIONS OF THE SECOND DEGREE. . a5 


yore, and also of ena is +q+=), and every 
quadratic equation must therefore have two roots. 


(182.) From the preceding discussion we deduce the follow- 
ing general 


RULE FOR THE SOLUTION OF A COMPLETE QUADRATIC EQUATION. 


1. Transpose all the known quantities to one side of the equa- 
tion, and all the terms involving the unknown quantity to the 
other side, and reduce the equation to the form ax*+bx=c. 

2. Divide each side of the equation by the coefficient of x’, and 
add to each member the square of half the coefficient of the first 
power of x. 

3. Hxtract the square root of both sides, and the equation will 
be reduced to one of the first degree, which may be solved in the 
usual manner. 


‘EXAMPLE l. 


Solve the equation z?—10x=— 16. 
Completing the square by adding to each member the square 
of half the coefficient of the second term, we have 
x’ —102+25=25—16=9. 
Extracting the root, xz—5= +3. 
Hence = 53. 
ape, zr=5+3=8, 
L=5—3=2. 
Thus x has two values, either 8 or 2. To verify them, sub- 
stitute in the original equation, and we shall have 
8’—10x8=— 16, 7. e., 64—80=— 16; 
also, 2?—10X2=—16,1.¢e, 4—20=—16, 
both of which are identical equations. 


EXAMPLE 2. 


Solve the equation 2+6r=—8, 
Completing the square, 7°+62+9=9—8=l. 
Extracting the root, «#+38=+1. 
Hence b=—Zt1. 
z=—3+1=—2, 
Ans. r=—3—1=—4, 
K 


‘s 


Ny 
a 
har 


146 detuntoNe OF THE SECOND DEGREE. 

Proof. (—2)’+6X—2=—8,i.e, 4—12=—8; 
also, (—4)’+6x —4=—8, 2. e., 16—24=—8. 

Hence x has two values, both negative. In verifying them, 
it is to be observed, that the square of —2 is +4,and —2 mul- 
tiplied by +6 gives —12. 


EXAMPLE 3. 


Solve the equation x’ -+62=27. 

Completing the square, 27+ 6z+9=27+9=86. 
Extracting the root, xz +3=-46. 

Hence x =—3+6=-+43, or —9. 


EXAMPLE 4, 
Solve the equation 7?—2r=24. 
Here x =1+5=-+6, or —4. 


EXAMPLE 5. 


Solve the equation x’—8zx=-— 18. 
Completing the square, z7—8zx+16=16—18=—2. 
Hence g = of — 2, 


Here both values of x are imaginary. 


EXAMPLE 6. 


Solve the equation x'—6x=—9. 

Completing the square, z’°—6z+9=9—9=0. 

Extracting the root, 2 —3=+0. 

Hence 2 =30. 

Here the two values of x are equal to each other. 

Ex. 7. Given 2z°+8z—20=70, to find the values of x. 

Ans. x=5, or —9. 

Ex. 8. Given 3x’—32+6=5}, to find the values of z. 

Ans. 2=%, Ora 

(183.) The Rule given on page 145 for solving a complete 
quadratic equation is applicable to all cases; nevertheless, a 
modification of this method is sometimes preferable. 

The object is to render the first member of the equation a 
perfect square. After the equation has been reduced to the 
form 

ax’*+bz=c, — 


EQUATIONS OF THE SECOND DEGREE. 147 


the square may be completed by multiplying the equation by 
four times the coefficient of x’, and adding to both sides the 
square of the coefficient of x. 
Thus the above equation multiplied by 4a becomes 
4a°x’ +4abx=4ac. 

Adding b* to both members, we have 

4a*x*+4abz+b’=4ac+b’. 

Extracting the square root, | 

2ax+b==+ v 4ac+0b’. 
Transposing 0b and dividing by 2a, 
—b+ V4ac+b' 
jidauo Viw 2a i 
which is the same result as would be obtained by the former 
Rule; but by this new method we have avoided the introduc- 
tion of fractions in completing the square. 

When the coefficient of z’ is unity, the above Rule becomes, 
Multiply the equation by four, and add to each member the 
square of the coefficient of x. 

Hither of these methods of completing the square may be 
practiced at pleasure; but the first method is to be preferred, ex- 
cept when its application would involve inconvenient fractions. 

Ex. 9. Given }2’—17+201=4232, to find the values of z. 

Ans. ~=", or —6}. 

Ex. 10. Given z’—x—40=170, to find the values of z. 

Ans. x=15, or —14. 





4 pe 


Ez. 11. Given 3z?+2z7—9="6, to find the values of 2... %. fe oY 


Ex. 12. Given 12’—12+72=8, to find the values of x... 


6z7—40 327—10 
Q—1 9— 





Ez. 13. Given 8x— =2, to find the values 


of z. 

We must first clear this equation of fractions, which is done 
by multiplying by the denominators ; we thus obtain 

—122°+ 602*— 272+ 122°— 5427+ 360—802—62?+232—10 

=407—8z’— 18. 

Here‘ the two terms containing 2° balance each other, and, 
uniting similar terms, we obtain 

8z*—124c=— 368. 


148 EQUATIONS OF THE SECOND DEGREE. 


Dividing by 8, we have 
x’ —15ixr=—46. 
Completing the square, 
31 225 
2 —15j2+(9 zit =~46+(* =) aa tf: 


3] 15 
Extracting the root, traqeae 


4 
ae Bes hi 
Hence ter Tate ge Lhd or 4. Ans. 
90 90 o7 
1 aie AI — : 
Eiz. 14. Given Tae aa —0, to find the values ae eS 


(184.) The preceding rules will enable us to sole noe aa 
quadratic equations, but all equations which can be reduced to 
the form 

n+ pxr"=¢ 5 
that is, all equations which contain only two powers of the un 
known quantity, and in which one of the exponents 1s double of 
the other. 

For if, in the above equation, we assume y=z", then y’=2”, 
and it becomes 

y tpy=4- 

Solving this according to the rule, we find 


maya Pac gah . 


Extracting the nth root of both sides, 
z= af Pak Figo ont ae gre 


EXAMPLE lI. 


Given z*—25z’?=— 144, to find the values of z. 
Assuming z’=y, the above becomes 


y?—25y=—144. 
Whence . y= 16, or 9. 
But, since 2’°=y, ea a/y. 
Therefore, pera / 1G, ar cf O) 


Thus z has four values, viz., +4, —4, +38, —3. 


ov 


EQUATIONS OF THE SECOND DEGREE. 149 


To verify these values: 
Ist value, (+4)*—25x (+4)’=—144, 7. e., 256—400=— 144. 
2d value, (—4)*‘—25x (—4)’=— 144, 1. e., 256—400=— 144, 
3d value, (+8)*—25x (+38)’=— 144, i.e, 81—225=—144., 
4th value, (—3)'—25x (—8)’=— 144, i. e., 81—225=—144. 


EXAMPLE 2. 
Given z*—7z’=8, to find the values of z. 
Assuming x’=y, we have 
y'—'ty=8. 
Whence y =8, or —1. 
Therefore, z==+ V8, or + V —1, the two last of which roots 
are imaginary. 


EXAMPLE 3. 
Given x°—2z°=48, to find the values of z. 
Assuming x°=y, the above becomes 
y?—2y=48. 
Whence y =8, or —6. 


And since «*=y, therefore z= Vy. 

Hence two of the roots of the above equation are 2 and— //6. 

This equation has four other roots, which can not be de- 
termined by this process. 


EXAMPLE 4. 


Given 22—7./x=99, to find the values of z. 
Assuming /z=y, this equation becomes 


2Qy?—71y=99. 

1] 

Whence y=9, or a 
And since /x=y, therefore z=y’. 
121 

Whence $= 81. OF cai 


Although the square root of 81 is generally ambiguous, and 
may be either +9 or —9, still, in verifying the preceding 
values, x can not be taken equal to —9, because 81 was ob- 
tained by multiplying +9 by itself. For a like reason, Vz 


11 Baa ty Pabst 
can not be taken equal to stk: A similar remark is applica- 


150 EQUATIONS OF THE SECOND DEGREE. 


ble to Evs. 13 and 14 on the next page, and also to Ez. 7, 
page 186. 


EXAMPLE DB. 


Given Vz+12+ Vx+12=6, to find the values of x. 
Assuming x+12=y, this equation becomes 
1 1 
) y’ +y*=6, 
which evidently belongs to the same class as the previous ex- 
amples. Completing the square, we shall have 


1 
y* =2, or —3. 
Raising both sides of the equation to the fourth power, 
y=16, or 81, 
Therefore, _ x or (y—12)=4, or 69. 
EXAMPLE 6. 


Given 2z7°+ V2z’?+1=11, to find the values of z. 
Adding 1 to each member of the equation, it becomes 
Q2°+1+ ¥227?+1=12. 


Assuming 2x°+1=y, then 
1 
yt+y?=12. 
Completing the square, we find 
BE 
y’ =3,.0or —4; 
that is, V 2x7°+1=3, or —4. 
Therefore, 27°+1=9, or 16, 
: 15 
and z°=A, o1 3° 


H =+2 NA fat Ace 
ence roa +9, ; 9? 3° 


It may be remarked, that in equations of this kind it is gen- 
erally unnecessary to substitute a new letter, 7, which has been 
done in the preceding examples simply for the sake of per- 
spicuity. | 

Ez. 7. Given z*+42’?=12, to find the values of z. 

Ans. c=+./2, or +V—6. 


\ 
EQUATIONS: OF THE SECOND DEGREE. 151 


Ex. 8. Given 2°—82z'°=518, to find the values of z. 


Ex 


Ans. x=3, or — V19. 


6 KS 
. 9. Given x°+2°=1756, to find the values of z. 


Ans. 7=243, or — °/ 28°, 


Ex. 10. Given 12°—12°=— 44, to find one value of z. 


Ex 


Ex 
of z. 


Ansi ce Lal 2. 


2 1 
11. Given 22°+3z°= 2, to find the values of z. 
Ans. x=1, or —8. 


. 12. Given z2*—122°+ 442?— 48z=9009, to find the values 


This equation may be expressed as follows: 


Ez. 


Ez. 


(x*—6z)?+8(2?— 6z) =9009. 
Ans. x=13, or —7, or 3 V —90. 


. 18. Given 17—1./r=223, to find the values of z. 


361 
Ans. x=49, or coi 


. 14. Given V10+2— V10+27=2, to find the values of z. 


Ans. x=6, or —9. 
15. Givem2*+202°*—10=59 to find the values of z. 


Ans. x= V3, or / —23. 
16. Given 3z2"—22"+3=11, to find the values of z. 


—_—_— 


Ans. c= V2, or V—4, 


. 17. Given z’—x./3=x—1,3, to find the values of z. 


J/3+3 PY es Feat | 
or 


9 . 
2 2 


Ans. 








Ez.18. Given V1+x—z*—2(1+2x—2")=}, to find the values 


of x. 


Ans. 1+1V41, or 11V11. 


(185.) We have seen that every equation of the second de- 
gree has two roots, or that there are two quantities which, when 
substituted for z in the original equation, will render the two 
members identical. In like manner, we shall find that every 
equation of the third degree has three roots ; an equation of the 
fourth degree has four roots; and, in general, an equation has 
as many roots as it has dimensions. 


152 PROBLEMS PRODUCING QUADRATIC EQUATIONS. 


Before determining the degree of an equation, it should be 
freed from fractions, from negative exponents, and from the 
radical signs which affect its unknown quantities. Examples 
4, 5, 13, and 14 are thus found to furnish equations of the sec- 
ond degree, while examples 6 and 18 furnish equations of the 
fourth degree. 

The above method of solving the equation 2”+pz"=q will 
not always give us all of the roots, and we must have recourse 
to different processes to discover the remaining roots. The 
subject will be resumed in Section XX. 


PROBLEMS PRODUCING QUADRATIC EQUATIONS. 


Prob. 1. It is required to find two numbers, such that their 
difference shall be 8, and their product 240. 
Let x = the least number. 
Then will +8 = the greater. 
And by the question z(7+8)=2?+8zr=240. 
Therefore, x=12, the less number, 
x+8=20, the greater. 
Proof. 20—12=8, the first condition. 
20x 12=240, the second condition. 
Prob. 2. The Receiving Reservoir at Yorkville is a rectan- 
gle, 60 rods longer than it is broad,.and its area is 5500 ) Aare 
rods. Required its length and breadth? 


Prob. 3. What two numbers are those whose autora is 
2a, and product b? 
Ans. ak Va?+b, and —a+ Va? +b. 
Prob. 4. It is required to divide the number 60 into two such 
parts that their product shall be 864. 
Let x = one of the parts. 
Then will 60—z = the other part. 
And by the question, «(60—2x)=60z7 —2z’=864. 
The parts are 36 and 24. Ans. 
Prob. 5. In a parcel which contains 52 coins of silver and 
copper, each silver coin is worth as many cents as there are 
copper coins, and each copper coin is worth as many cents as 
there are silver coins, and the whole are worth two dollars. 
How many are there of each? 


S 5D vedo — WB: 


Cc) 
Fb 21% 


2 Ae DI Wtrer pele Pe & OP vg a a 62017 


¢ 


PROBLEMS PRODUCING QUADRATIC EQUATIONS. 153 


Prob. 6. What two numbers are those whose sum is 2a, and 
product b? 


Ans. a+ Va'?—b, and a— Va’?—b. 
Prob. 7. There is a number consisting of two digits whose 


sum is 10, and the sum of their squares is 58. Required the 
number. 


Let x = the first digit. : 
‘Then will 10—a2 = the second digit. 
And x’+(10—2z)’=22°—20x+100=58. 


That is, 2?—10r—=—21, 


x’ —10x2+25=4, 
SHAR esl; OF 3! 
Hence the number is 73, or 37. 


The two values of x are the required digits whose sum is 
10. It will be observed that we put z to represent the first 
digit, whereas we find it may equal the second as well as the 
first. ‘The reason is, that we have here imposed a condition 
which does not enter into the equation. If z represent either 
of the required digits, then 10—z will represent the other, and 
hence the values of x found by solving the equation should 
give both digits. Beginners are very apt thus, in the state- 
ment of a problem, to impose conditions which do not appear 
in the equation. 

The preceding example, and all others of the same class, 
may be solved without completing the square. Thus, 

Let z represent the half difference of the two digits. 

Then, according to the principle on page 67, 
5+<2 will represent the greater of the two digits, 


5-2 s the less a 

The square of 54+2 is 25+10r+ 2’, 

ig 5—a2 25-1074 2’, 

The sum is 50 +227, which, according to the 
problem, = 58, 

Hence Bo *=3 1.8, 
or haw A, 
and peg 

Therefore, 5+2 =7, the greater digit, 


5—zx =3, the less digit. 


ol - 
154 PROBLEMS PRODUCING QUADRATIC EQUATIONS. 


Prob. 8. Find two numbers such that the product of their 
sum and difference may be 5, and the product of the sum of 
their squares and the difference of their squares ey) be 65. 


Prob. 9. Find two numbers such that the prorat of their 
sum and difference may be a, and the product of the sum,of 
their squares and the difference of their squares may be ma. 


Ans. \/ ae 


Prob. 10.. A laborer dug two trenches, whose united length 
was 26 yards, for 356 shillings, and the digging. of each of 
them cost as many shillings per yard as there were yards in 
its length. What was the length of each? 

Ans. 10, or 16 yards. 





Prob. 11. What two numbers are those whose sum is 2a, 
and the sum of their squares is 2b ? 
Ans. a+ Vb—a’, and a— Vb—a’. 


Prob. 12. A farmer bought a number of sheep for 80 dollars, 
and if he had bought four more for the same money, he would 
have paid one dollar less for each. How many did he buy ? 


Let z represent the number of sheep. 


Then will a be the price of each. 


And 4 would be the price of each, if he had bought four 


more for the same money. 
But by the question we have 
80 - 80 ie Oorey 
a at+4 
Solving this equation, we obtain 
c= 16. Ans: 
Prob. 13. A person bought a number of articles for a dol- 
lars. Ifhe had bought 2b more for the same money, he would 
have paid c dollars less for each. How many did he buy? 


Ans. —b+ / a 


Prob. 14. It is required to find three numbers such that the 


F i 


PROBLEMS PRODUCING QUADRATIC EQUATIONS. 155 


product of the first and second may be 15, the product of the 
first and third 21, and the sum of the squares of the second and 
third 74. 
Ans. 8, Spands7; 
Prob. 15. It is required to find. three numbers such that the 
product of the first and second may be a, the product of the 
first and third 6, and the sum of the squares of the second and 


third c. 
a+b? V/ c V c 
Anas a ep OV aR 


Prob. 16. The sum of two numbers is 16, and the sum of 
their cubes 1072. What are those numbers ? 
Ans. 7 and 9. 


Prob. 17. The sum of two numbers is 2a, and the sum of 
their cubes is 2b. What are the numbers? 


Ans. a+ VJ 


mE and a— al 
Prob. 18. Two magnets, whose powers of attraction are as 
4 to 9, are placed at a distance of 20 inches from each other. 
It is required to find, on the line which joins their centers, the 
point where a needle would be equally attracted by both, ad- 
mitting that the intensity of magnetic attraction varies inverse- 

ly as the square of the distance. 
Fide: 8 inches from the weakest magnet, 
or —40 inches from the weakest magnet. 











Prob. 19. Two magnets, whose powers are as m to n, are 
placed at a distance of a feet from each other. It is required 
to find, on the line which joins their centers, the point which is 
equally attracted by both. 
| a/m 
Vm /n 


Ans 
a/n 
The distance from the magnet n is — vf 


The distance from the magnet m is 





Jm+ /n’ 


Prob. 20. A set out from C toward D, and traveled 6 miles 
an hour. After he had gone 45 miles, B set out from D to- 
ward C, and went every hour 3; of the entire distance; and 
after he had traveled as many hours as he went miles in one 


156 ' QUADRATIC EQUATIONS 


hour, he met A. Required the distance between the places © 
and D. 
Ans. Either 100 miles, or 180 miles, 


Prob. 21. A set out from C toward D, and traveled a miles 
per hour. After he had gone 0 miles, B set out from D toward 
C, and went every hour ;th of the entire distance; and after 
he had traveled as many hours as he went miles in one hour, 
he met A. Required the distance between the places C and D. 


Ans. (tae) (ae 


Prob. 22. By selling my horse for 24 dollars, I lose as much 
per cent. as the horse cost me. What was the first cost of 
the horse ? 





Ans. 40, or 60 dollars. 


QUADRATIC EQUATIONS CONTAINING TWO UNKNOWN QUAN. 
TITIES. 

(186.) An equation containing two unknown quantities is 
said to be of the second degree when it involves terms in which 
the sum of the exponents of the unknown quantities is equal to 2, 
but never exceeds 2. Thus, 

ax’ —4x2+y'=25, 
and Tzy—4z+y =40, 
are equations of the second degree. 

The solution of two equations of the second degree contain- 
ing two unknown quantities, generally involves the solution of 
an equation of the fourth degree containing one unknown quan- 
tity. Hence the principles hitherto established are not sufh- 
cient to enable us to solve all equations of this description. 
Yet there are particular cases in which they may be reduced 
either to pure or affected quadratics, and the roots determined 
in the ordinary manner. 

(187.) When one of the equations is a simple equation, it is 
generally best to find an expression for the value of one of the 
unknown quantities from the simple equation, and substitute 
this value in the place of its equal in the other equation. The 
resulting equation will be of the second degree, and may be 
solved by the ordinary rules. 


CONTAINING TWO UNKNOWN QUANTITIES. 157 


Ex. 1. Given 2°+3ry— y’=23 ) to find the values of zx © 
z+2y= 7 and y. 
From the second equation, we find 
x =T—2y. 
Whence x’? =49—28y+4y’. 
And, substituting this value in the first equation, we have 
49—28y+4y*+2ly—6y’*—y’=23, 
a common quadratic equation, which may be solved in the 
usual manner. 
Ans. x=3, and y=2. 
Ez. 2. Given 2z°+ zxry—5y’=20 to find the values of z 
2% —3y= | and y. 
Ans. x=5, y=3. 
: 10z+ 
Maas S4Yen sn td 1 find the values of x and y. 
9y—9zx=18 
Ans. r=2,/'y=4. 
(188.) When the same algebraic expression is involved to 
different powers, it is sometimes best to regard this expression 
as the unknown quantity. | 
Ez. 4. Given z’+2zy+y? +22=120—2y ) to find the val- 
xy—y" =8 ues of x and y. 
Here the first equation may be put under the form 
(v+y)’+2(a+y)=120, 
where z+y may be regarded as a single quantity, and, by 
completing the square, we shall find its value to be 
either 10, or —12. 
Proceeding now as in the last Article, we shall find 
x=6, or 9, or —9> V5, 
y=4, or 1, or —8+ 75. 
Ex. 5. Given 4ry=96—2’y’ 
x+y=6 
Here we may regard zy as the unknown quantity, and we 
shall find its value from the first equation to be 
either 8, or —12, 


Proceeding again as in the former Article, we shall find 


to find the values of z and y. 


4 
4 


"> , 
gts : “ 7 
158 % QUADRATIC EQUATIONS ~ ta 
et? 2 x=2, or 4, or 8+ y2l, 
pe y=4, or 2, or 3+ 21. 
wie Az _ 
6. Given —+—=— 
aes Re y at y to find the values of x and y. 


xL—Yy= ay 
a 
Here i may be treated as the unknown Cer and we 


shall find its value to be 


either gy  — 3° | 
From which we find 
17 
‘ zZ=5, OY io’ 
3 
y=, Or 10° 





(189.) When the sum of the dimensions of the unknown 
quantities is the same in every term of the two equations, it is 
sometimes best to substitute for one of the unknown quantities 
the product of the other by a third unknown quantity. 

- Ex. 7. Given z?+2y =56 
zy +2y’=60 

Here, if we assume z=vy, we shall have 

vy’ +vy’=56, 
vy’ +2y'=60. 
From the first of these equations, 
=. 16 
Lee 
_ 60 
es v +2? 
60 56 
v+2 vi+v' 
From which, after completing the square, we obtain 
4 ci 
v= 5) or pa 

Substituting either of these values in one of the preceding 
expressions for y’, we shall obtain the values of y; and since 
cmey, we may easily obtain the values of z. 

x=-b14, OF =E4yn, 
Ans. y==+10, or =3/2. 


to find the values of z and y. 


and from the second, 


therefore, 


a. 


| loi ine all = a 7 - Taal he 
i 
; . 
: » ‘ . i's 
9 . 4 te: Hr 
CONTAINING TWO UNKNOWN QUANTITIES. 159 7 
a, er af 


Ex. 8. Given Qx°+3cy+ y=20 to find the values of z 
/ 52* +4y?=41 and y. } 
If we assume x=vy, we shall find 
ice. WEL ‘s 
Ar or 9” 
whence, as before, we shall obtain 


( 


aap 
| Tiel, Of ys 
V¥21 


Ans. ae 
y=+3, on 
V¥21 


meen etry to find the values of x and y. 


ry—y =12 
If we assume x=vy, we shall find 
7 1] 
UTig or 3? 


whence, as before, 


[ ons +1] 
r= Chan $ 
2 
Ans. A 
—+4 =3 
een 9 or Ty 


(190.) When the unknown quantities in each equation are 
similarly involved, it is sometimes best to substitute for the 
unknown quantities the sum.and difference of two other quan- 
tities, or the sum and product of two other quantities. 


en ee 
Ez. 10. Given y an a 8 c find the values of z and y. 


t+y =12 
Here let us assume 
c=z+uv, 
Yy=27—v. 
Then, by adding these two equations together, we shall have 
L+Yy=2z2=12, or z=6; 
that is, x=6-+0, and y=6—v. 
But, from the first equation, we find i. 
xz’ +y'=18zry. 


EY 
“a 


160 "+ QUADRATIC EQUATIONS 


Substituting the preceding values of 2 and y in this equa- 
tion, and reducing, we obtain 


432+ 36v°=648— 180’. 


Whence U—-2, 
Therefore, e=A, or 8, 
and y=8, or 4. 
Ex. 11. Given x°+y°=3368 ; 
q yen aye to find the values of x and y. 
z+ty= 8 
Ava =o, OY 0, 
y=5, or 3. 





Ex. 12. Given z* +y° =341 
aty+ny'=880 to find the values of x and y. 
2=), Ora 


Ans. | he 





PROBLEMS. 


1. Divide the number 100 into two such parts, that the sum 
of their square roots may be 14. 


Ans. 64 and 36. 


2. Divide the number a into two such parts, that the sum 
of their square roots may be 0. 


Que) / eee 
Oe = i: my 
Ans 5 5 V 2a b 


3. The sum of two numbers is 8, and the sum of their fourth 
powers is 706. What are the numbers? 
Ans. 3 and 5. 


4. The sum of two numbers is 2a, and the sum of their 
fourth powers is 2b. What are the numbers ? 
Ans. atV —8a'?+ V8a*+0. 
5. The sum of two numbers is 6, and the sum of their fifth 
powers is 1056. What are the numbers ? 
Ans. 2 and 4. 


6. The sum of two numbers is 2a, and the sum of their fifth 
powers is b. What are the numbers ? 


pad Ree 
Ans. at i a’, 


CONTAINING TWO UNKNOWN QUANTITIES. 161 


7. What two numbers are those whose product is 120 ; and 
if the greater be increased by 8 and the less by 5, the product 
of the two numbers thus obtained shall be 300? 

Ans. 12 and 10, or 16 and 7.5. 

8. What two numbers are those whose product is a; and 
if the greater be increased by 6 and the less by c, the product 
of the two numbers thus obtained shall be d? 


Ans. Mahe 2 se and = Jw 


scat 





' where m= 


9. Find two numbers such that their sum, their product, 
and the difference of their squares may be all equal to one an- 
other. 


Ans. “+ va and 5+ Vy 
that is, 2.618, and 1.618, nearly. 
10. Divide the number 100 into two such parts, that their 
product may be equal to the difference of their squares. 
Ans. 38.197, and 61.803. 


11. Divide the number a into two such parts, that their prod- 
uct may be equal to the difference of their squares. 


8ata/J/5 —amza./d5 
Ans. 5 ae and mu (sie 








DISCUSSION OF THE GENERAL EQUATION OF THE SECOND 
DEGREE. 


(191.) We have seen, Art. 181, that every equation of the 
second degree may be reduced to the form 
xz’+pr=q, | 
where p and q represent known quantities, either positive or 
negative, integral or fractional. 
The value of z in this equation is 


either oon Ta? 


p 2 
or w=—P_y/ 94H. 
L 


, 7 


162 DISCUSSION OF THE. 


And, since these values necessarily result from the general 
equation, we infer, © 


PROPERTY I. 


Every equation of the second degree has two roots, and only 
two. 

A root of an equation is such a number as, being substituted 
for the unknown quantity, will satisfy the equation. 

This principle has been often exemplified in the preceding 
pages. ‘Two values have uniformly been found for z, although 
both values may not be applicable to the problem which fur- 
nishes the equation. This property will be found demon- 
strated in a general manner in Art. 294. 


(192.) If we multiply 
ey on 
tts O+ A 0, 


Bey gt 
by ttoty qt 70, 
-we shall obtain x’ +-pz—q=0, 


which was the equation originally proposed. 
Hence, 


PROPERTY II. 


Every equation of the second degree, whose roots are a and b, 
may be resolved into the two factors x—a and x—b. 
Ex. 1. Thus the equation 
x*+'1 02 4-PE==0, 
x—8=—0, 
r—2=—0, 
where 8 and 2 are the roots of the given equation. 


It is also obvious that if a is a root of an equation of the sec- 
ond degree, this equation must be divisible by z—a. Thus 
the preceding equation is divisible by z—8, giving the quotient 
L—2. ees 


may be resolved into the factors 


Ex. 2. The roots of the equation 





ro ah am ’- «A 


GENERAL EQUATION OF THE SECOND DEGREE. 163 


z’?+6z2+8=0, 
are —2 and —4. Resolve it into its factors. 
Ex. 3. The roots of the equation 
x’°+6x2—27=0, 
are +3 and —9. Resolve it into its factors. 
Ez. 4. The roots of the equation 
x’*—2xr—24=0, 
are +6 and —4. Resolve it into its factors. 


(193.) If we add together the two values of x in the gen- 
eral equation of the second degree, the radical parts having 
opposite signs disappear, and we obtain 


ya a 


Tomes = —p. 
Hence, 


PROPERTY TIL. 


The algebraic sum of the two roots is equal to the coefficient 
of the second term of the equation, taken with a contrary sign. 
Thus, in Hz. 1, page 145, 
x*—10x#=— 16, , 
the two roots are 8 and 2, whose sum is +10, the coefficient 
of x taken with a contrary sign. 
In the equation 


z’+62=——8, 

the two roots are —2 and —4. 
In the equation 

xz’°+16xz2=—60, 

the two roots are —6 and —10. 


If the two roots are equal numerically, but have opposite 
signs, their sum is zero, and the second term of the equation 
vanishes. ‘Thus the two roots of the equation 2°=16, are +4 
and —4, whose sum is zero. This equation may be written — 

xv’+0rz=16. 

(194.) If we multiply together the two values of x (observ- 
ing that the product of the sum and difference of two quan- 
tities is equal to the difference of their squares), we obtain 


, oo 
oa 


164 DISCUSSION OF THE 


eh iw is aaa 
4 (q+4) = q: 


Hence, 


PROPERTY IV. 


The product of the two roots is equal to the second member of 
the equation, taken with a contrary sign. 


Thus, in the equation 
xz’ —10z7=— 16, 
the product of the two roots 8 and 2 is +16, which is equal to 
the second member of the equation taken with a contrary sign. 
So, also, in the equation | 
r°+627=27, 
whose two roots are +3 and —9, their product is —27. 


The two last properties enable us readily to form an equa- 
tion when its roots are known. 


oe 


Ex. 1. Let it be required to form the equation whose roots 
are 2 and 8. 


According to Property IL, the coefficient of the second 
term of the equation must be —10; and, from Property IV., 
the second member of the equation must be —16. Hence the 
equation is 

| a’—10c=—16. 

Ex. 2. Form the equation whose roots are 3 and 5. 

Ex. 3. Form the equation whose roots are —4 and —‘. 

ix. 4. Form the equation whose roots are 5 and —9. 

fix. 5. Form the equation whose roots are —6 and +11. 


REAL AND IMAGINARY VALUES OF THE UNKNOWN QUANTITY. 


(195.) The values of x in the general equation of the second 


degree are 
2 ANTE: 


Values of the unknown quantity which are not imaginary, 
are, for the sake of distinction, called veal. 


GENERAL EQUATION OF THE SECOND DEGREE. 165 
2 
Since S being a square, is positive for all real values of p, 


2 


it follows that the expression qt7 can only be rendered neg- 


ative by the sign of q. 
When q is positive, or when g is negative and numerically 


2 


less than ‘, then will q+ be positive, and, consequently, 





V/ q+4 will be real. This happens in nearly all the preced- 
ing examples. 


2 
When q is negative, and numerically greater than mo then 


qe will be negative, and, consequently, Ve q+4 will be im- 
aginary. This happens in Ez. 5, page 146. 


CASE I. 
When J. iar 
Qie ra real. 
1. When, in the equation z’+pzr=q, p is negative, and 


is numerically greater than V/ q+4, both values of x will be 


real and positive. 
This happens in the equation 
x’ —b6z=—8, 
whose two roots are 4 and 2, 
Also in the equation - 
z?—10z=—16, 
whose two roots are 8 and 2. 


2. When p is positive, and ‘ is numerically greater than — 





J q+ both values of x will be real and negative. 


This happens in the equation 


we 


— < 4 7 = > i% os, Sy) eo a 
- , 


ra, 
166 DISCUSSION OF THE 


xv’+6zr2=—8, 
whose two roots are —2 and —4. 
Also in the equation 
z’*+16z=— 60, 
whose two roots are —6 and —10. 


3. When - is numerically less than / q+4, both values of 
x will be real, the one positive and the other negative. 
This happens in the equation 
xz’*+62=27, 
whose roots are +3 and —9. 
Also in the equation 
x? —Qr=24, 
whose roots are +6 and —4. 


CASE II. 


(196.) When J q+e is imaginary. 


In this case, both values of x are imaginary. 
This happens in the equation 
x’*—8x=— 18, 


whose roots are 4+ ¥ —2, 

We will now prove that in this case the conditions of the 
question are incompatible with each other, and therefore the 
values of x ought to be imaginary. The demonstration de- 
pends upon the following principle: 

The greatest product which can be obtained by dividing a 
number into two parts and multiplying them together, is the 
square of half that number. 


Let p = the given number, 
and d= the difference of the parts. 
Then, from page 67, Dts = the greater part, 


ae fee 
ey the less part, 


GENERAL EQUATION OF THE SECOND DEGREE. 167 


2 2 


and — = the product of the parts. 


Now, since p is a given quantity, it is plain that this expres- 
2 


sion will be the greatest possible when d=0; that is, f is the 


greatest product, which is the square of r, half the given 


number. 
For example, let 12 be the number to be divided. 
We have 12=1-+] 1); and 11X1=11., 
12=2+10; and 10x2=20. 
12=34+ 9; and 9x3=27. 
12=4+ 8; and 8xX4=82. 
12=5+ 7; and 7X5=35. 
12=6+ 6; and 6X6=36. 

We here see that the smaller the difference of the two parts, 
the greater is their product; and this product is fit when 
the two parts are equal. 

Now, in the equation 


x’ —pzr=—4q, 
p is the sum of the two roots, and q is their product. There- 
fore q can never be greater than “ 


If, then, any problem furnishes an equation in which q is 


2 


negative, and greater than ‘, we infer that the conditions of 


the question are incompatible with each other. 

Thus, in the example 

xz’—b62= oe 10, 

f=9, which is numerically less than g. The equation re- 
quires us to divide the number 6 into two parts whose product 
shall be 10, which is an impossibility ; and, accordingly, in 
solving the equation, we obtain imaginary values for z. 

Hence an imaginary root indicates an absurdity in the pro- 
posed question which furnished the equation. 

Suppose it is required to divide 8 into two such parts that 
their product shall be 18. 


kis be « akg Bak [ a Ve 


on 
oh 


4 


168 “DISCUSSION OF PARTICULAR PROBLEMS. 
Let x = one of the parts, 
and 8—z = the other. 
Then, by the conditions, 
x(8—zx)=18. 
Whence x’ —8x=— 18. 


This equation, solved by the usual method, gives 

a=4+ ¥ —2, an imaginary expression. 

Hence we infer that it is impossible to find two numbers 
whose sum is 8, and product 18. This is obvious from the 
Proposition above demonstrated, from which it appears that 
16 is the greatest product which can be obtained by dividing 
8 into two parts, and multiplying them together. 


197.) When gq is negative, and numerically equal to io the 
(197.) q is neg y eq 7 


radical part of both values of x becomes zero, and both values 


of z reduce to —£. The two roots are then said to be equal. 


Thus, in the equation 
xz’°—6z=-—9, 
the two roots are 8 and 8. 


We say that in this case the equation has two roots, because 
it is the product of the two factors, z—3=0, and z—3=0. 


DISCUSSION OF PARTICULAR PROBLEMS. 


(198.) In discussing particular problems which involve equa- 
tions of the second degree, we meet with all the different cases 
which are presented by equations of the first degree, and some 
peculiarities besides. We may therefore have, 


1. Positive values of z. 
2. Negative values. 

0 
3. Values of the form of z 


4, Values of the form of * 


5. Values of the form of > 


‘ All these different cases are presented by Problem 19, 


y’ Ww 
DISCUSSION OF PARTICULAR PROBLEMS. @ 169 


page 155, when we make different suppositions upon the values 
of a, m, and n; but we need not dwell upon them here. 

The peculiarities exhibited by equations of the second de- 
gree are, * 

6. Double values of z. 

7. Imaginary values. 

We will consider the last two cases. 

(199.) Double values of the unknown quantity. 

We have seen that every equation of the second degree has 
two roots. Sometimes both of these values are applicable to 
the problem which furnishes the equation. Thus, in Problem 
20, page 155, we obtain either 100 or 180 miles for the dis- 
tance between the places C and D. 

C : D 
a Reuteiets O SAN. ealte! it i Seas 

Let E represent the situation of A when B sets out on his 
journey. Then, if we suppose CD equals 100 miles, ED will 
equal 55 miles, of which A will travel 30 miles (being 6 miles 
an hour for 5 hours), and B will travel 25 miles (being 5 miles 
an hour for 5 hours). 

If we suppose CD equals 180 miles, ED will equal 135 miles, 
of which A will travel 54 miles (being 6 miles an hour for 9 
hours), and B will travel 81 miles (being 9 miles an hour for 
9 hours). 

This problem, therefore, admits of two positive answers, 
both equally applicable to the question. 

Problem 22, page 156, is of the same kind; and another 
will be found on page 193. 

In Problem 18, page 155, one of the values of z is positive, 
and the other negative. 


B 

3 EE ON taal Sed eerie 
Let the weakest magnet be placed.at A, and the strongest 
at B; then C will represent the situation of a needle equally 
attracted by both magnets. According to the first value, the 
distance AC=8 inches, and CB=12. Now at the distance of 
8 inches, the attraction of the weakest magnet will be repre- 


4 . : : 
sented by gi? and at the distance of 12 inches, the attraction 


¥ : 


iy wi” e 
170 ® DISCUSSION OF PARTICULAR PROBLEMS. 


9 
of the other magnet will be represented by ie and these two 


powers are equal; for 
49 
SOR 
But there is another point, C’, which equally satisfies the 
conditions of the question; and this point is 40 inches to the 
left of A, and therefore 60 inches to the left of B; for 
4.09 | 
40° 60" 
(200.) Imaginary values of the unknown quantity. 
We have seen that an imaginary root indicates an absurdity 
in the proposed question which furnished the equation. 
In several of the preceding problems, the values of x be 
come imaginary in particular cases. 
When will the values of z in Problem 6, page 153, be im- 
aginary ? 
Ans. When b>a’. 
What is the absurdity involved in this supposition ? 
Ans. It is absurd to suppose that the product of two num- 
bers can be greater than the square of half their sum. | 
When will the values of z in Problem 11, page 154, be imag- 
inary ? 
Ans. When a’?>b; or (2a)’>4b. 
What is the absurdity of this supposition ? 
Ans. The square of the sum of two numbers can not be 
greater than twice the sum of their squares. 


When will the values of z in Problem 17, page 155, be im- 
aginary ? 
Ans. When a°>b; or (2a)*>8b. 
What is thesabsurdity of this supposition ? 
Ans. The cube of the sum of two numbers can not be 
greater than four times the sum of their cubes. 


When will the values of z in Problem 4, page 140, be im- 
aginary, and what is the absurdity of this supposition ? 


SECTION XIII. 


RATIO AND PROPORTION. 


(201.) NumBers may be compared in two ways: either by their 
difference, or by their quotient. We may inquire how much one 
quantity is greater than another; or, how many times the one con- 
tains the other. One is called Arithmetical, and the other Geo- 
metrical Ratio. 

The difference between two numbers 1s called their Jrithmetical 
Ratio. Thus, the arithmetical ratio of 9 to 7 is 9 —7, or 2; and 
if a and } designate two numbers, their arithmetical ratio is rep- 
resented by a — 8. 

Numbers are more generally compared by means of quotients ; 
that is, by inquiring how many times one number contains another. 
The quotient of one number divided by another is called their 
Geometrical Ratio. The term Ratio, when used without any quali- 
fication, is always understood to signify a geometrical ratio, and 
we shall confine our attention to ratios of this description. 


(202.) By the ratio of two numbers, then, we mean the quotient 
which arises from dividing one of these numbers by the other. 


Thus, the ratio of 12 to 4 is represented by = or 3; 
The ratio of 5 to 2 is = or 2.5. 


The ratio of 1 to 3 is ‘ or .333. 


We here perceive that the value of a ratio cannot always be 
expressed exactly ; but, by taking a sufficient number of terms of 


172 RATIO AND PROPORTION. 


the decimal, we can approach as nearly as we please to the true 
value. 

If a and 6 designate two numbers, the ratio of a to d is the 
quotient arising from dividing @ by }, and may be represented by 
a 
* 
of the ratio, the last term, 0, is called the consequent of the ratio. 

Hence it appears that the theory of ratios is identified with the 
theory of fractions, and a ratio may be considered as a fraction 
whose numerator is the antecedent, and whose ; enominator is the con- 
sequent. 


writing them a: 6, or The first term, a, is called the antecedent 


(203.) When the antecedent of a ratio is greater than the con- 


Rr : 4 5-1 
sequent, the ratio is called a ratio of greater inequality ; as, 3 sia 


When the antecedent is less than the consequent, it is called a 


ratio of less inequality ; as, _ é. When the antecedent and con- 


8 
sequent are equal, it is called a ratio of equality ; as, 3) 9° It is 


plain that a ratio of equality may always be represented by 
unity. 


(204.) When the corresponding terms of two or more simple 
ratios are multiplied together, the ratios are said to be compounded. 


a ; ae 
Thus, the ratio of 3 compounded with the ratio of —, becomes 


ac 
bd’ 

When a ratio is compounded with itself, the result is called a 
duplicate ratio. ‘Thus, the duplicate ratio of 5 is . ; the duplicate 
a. a 
(ae i 
A ratio compounded of three equal ratios 1s called a traplicate 


ratio of 


ratio. Thus, the triplicate ratio of is Se the triplicate ratio of 


a re 
bay, 4OY 
The ratio of the square roots of two quantities is called a sué- 


RATIO AND PROPORTION. 173 


duplicate ratio. ‘Thus, the subduplicate ratio of : is a the sub- 
/a 


a 
duplicate ratio of — is —. 
p b Jb 


The ratio of the cube roots of two quantities is called a subtrip- 


; Are 8.5 2 
licate ratio. ‘Thus, the subtriplicate ratio of aq 18 3 the subtrip- 


Ya 


. ° a . 
licate ratio of — is ——. 
Cia 


(205.) If the terms of a ratio are both multiplied, or both divided by 
the same quantity, the value of the ratio remains unchanged. 


gory 
The ratio of a to d is represented by the fraction p and the 


value of a fraction is not changed if we multiply or divide both 
numerator and denominator by the same quantity. Thus, 


a 
ann man ns 
See tiluh nos 

n 


S| o 


a 
or, 2 Ona? mb —s— 
n 


(206.) Ratios are compared with each other by reducing the 
fractions which represent them to a common denominator. 

In order to ascertain whether the ratio of 2 to 7 is greater or 
less than that of 3 to 8, we represent these ratios by the fractions 
2 3 : 

7 and 9° and reduce them to a common denominator. They thus 
become 

16 oad 21 

56 56’ 
and, since the latter of these is the greatest, we infer that the 
ratio of 2 to 7 is less than the ratio of 3 to 8. 


(207.) A ratio of greater inequality is diminished, and a ratio of 
less inequality 1s increased, by adding the same quantity to both 
terms. 


174 RATIO AND PROPORTION. 
3 3-1 4, 
Thus, 5 > oEP or 55 
2 2+1 "4 
7) Sra 


a 

b 

‘and let x be added to each of its terms. The two ratios will 
then be 


To prove the proposition generally, let — represent any ratio, 


a d a+ 2x 
ke b+ 2 
which, reduced to a common denominator, become 
ab + an ab + bx 
b(b + wy b(b+ «) 


A OT Se ; ; 
If a > b, that is, if 5 is a ratio of greater inequality, then, since 
ax is greater than ba, the first of these fractions is greater than 


Gr, hear se 
the second, and therefore 7 8 diminished by the addition of the 
same quantity to each of its terms, 

a 
b 
ax is less than ba, the first of the above fractions is less than the 


But if a < 4, that is, if 


is a ratio of less inequality, then, since 


eb Me 
second, and therefore 7 is increased by the addition of the same 


quantity to each of its terms. 


(208.) If, in a series of ratios, the consequent of each is the 
antecedent of the following ratio, then the ratio of the first antece- 
dent to the last consequent is equal to that which is compounded of all 
the intervening ratios. 

Let the proposed ratios be 


Compounding them by Art. 204, we obtain 


abcde 
wie 


a. 


om 


eel 


RATIO AND PROPORTION. 175 


which, by dividing by dcde, reduces to 


SJ a 


PROPORTION. 


(209.) Proportion is an equality of ratios. 

Thus, if a, 6, c, d are four quantities, such that a, when divided 
by 6, gives the same quotient as c when divided by d, then a, b, c, d 
are called proportionals, and we say that a is tobascis tod; and 
this is expressed by writing them thus: 


or Aci 0-71, 
ac 
Bi Pau 


or 


In ordinary language, the terms ratio and proportion are con- 
founded with each other. ‘Thus, two quantities are said to be in 
the proportion of 3 to 5, instead of the ratio of 3to 5. A ratio 
subsists between ¢wo quantities, a proportion only between four. 
Ratio is the quotient arising from dividing one quantity by another ; 
two equal ratios form a proportion. 


(210.) In the proportion 
a:b::ce:d, 


a, b,c, dare called the zerms of the proportion. The first and last 
terms are called the extremes, the second and third the means. 
The first term is called the first antecedent, the second term the 
first consequent, the third term the second antecedent, and the fourth 
term the second consequent. 


(211.) When the second and third terms of a proportion are 
identical, this quantity is called a mean proportional between the 
other two. Thus, if we have three quantities, a, 5, c, such that 


176 RATIO AND PROPORTION. 


Reteab $100.2), 
then d is called a mean proportional between a and ¢, and c is: called 
a third proportional to a and b. 
If, in a series of proportional magnitudes, each consequent is 
identical with the next antecedent, these quantities are said to be 
in continued proportion. ‘Thus, if we have a, , c, d, e, f, such that 


a2 02: pews cl dc: a se spies 


or ret eee he 


the quantities a, b, c, d, e, f are in continued proportion. 


(212.) If four quantities are proportional, the product of the ex- 
tremes is equal to the product of the means. 


Let W102 Ce a, 
ey 
or 7 =F 


Multiplying these equals by bd, the expression becomes 


abd bus bcd 
re Wei Re 
or ad = be. 
Thus, if B se WORT, 
then 8x 12—4~x 9. 


(213.) Conversely, if the product of two quantities is equal to 
the product of two others, the first two quantities may be made the 
extremes, and the other two the means of a proportion. 


Let ad = be. 


Dividing these equals by bd, the expression becomes 


“ine Cayo 
Fommedil ald 

that is, Peo se wore C.. so 0, 
Thus, if Bp i Died: Ve 
then D Samet Ore 12. 


or Dl Bie 813 sede 


RATIO AND PROPORTION. 177 


(214.) This proposition is called the ¢est of proportions, and 
any change may be made in the form of a proportion which is 
consistent with the application of this test. In order, then, to 
decide whether four quantities are proportional, we compare the 
product of the extremes with the product of the means. 

Thus, to determine whether 5, 6, 7, 8 are proportional, we mul- 
tiply 5 by 8, and obtain 40. Multiplying 6 by 7, we obtain 42. As 
these two products are not equal, we conclude that the numbers 
5, 6, 7, 8 are not proportional. 

Again, take the numbers 5, 6, 10,12. The product of 5 by 12 
is 60, and the product of 6 by 10 is also 60. Hence the numbers 
are proportional ; that is, 


5.2 Orsi etOe 12. 


(215.) If three quantities are in continued proportion, ¢he prod- 
uct of the extremes is equal to the square of the mean. 


If a: ber? bis c. 
Then, by Art. 212, ac = bb = 8°. 


Conversely, if the product of two quantities is equal to the 
square of a third, the last quantity will bea mean proportional between 
the other two. 


Thus, let ac = 8. 


Dividing these equals by bc, we obtain 


a b 
1b. ror 
or te BUS bist 
Thus, if 4: Gis2 Ga 4, 
then 4'x 9 = 6’, 
And conversely, if dx, 9, 2x63, 


then 6 is a mean proportional between 4 and 9. 


EXAMPLES. 


1. Given the first three terms of a proportion, 24, 15, and 40, to 
find the fourth term. 


2. Given the first three terms = a proportion, 3ab°, 4070, and 


9a°b, to find the fourth term. 
M 


178 RATIO AND PROPORTION. 


3. Given the last three terms of a proportion, 4a°)°, 30°’, and 
2a°b, to find the first term. 

4. Given the first, second, and fourth terms of a proportion, 
by*, Tx*y*, and 21x°y, to find the third term. 

5. Given the first, third, and fourth terms of a proportion, 1, 
a —b, and a’ — b’, to find the second term. 


(216.) Ratios that are equal to the same ratio are equal to each other. 


Let ators acy, ovaries 
and Gidiimigg thena:b:erd 


For, from the first of these proportions, 


a_@ 
by 
And from the second, 
c ft 
dy 
Gaue 
Therefore, ,=7 
or (os OE Rae: 


(217.) If four quantities are proportional, they will be propor- 
tional by alternation ; that is, the first will have the same ratio to the 
third that the second has to the fourth. 


Let Bt Betis ces f 

Then, also, at Bro pr iP | 

For, applying the test of Art. 214, we have in each case 
ad — be. 


(218.) If four quantities are proportional, they will be propor- 
tional by znversion ; that is, the second will have to the first the same 
ratio that the fourth has to the third. 

Let OG ssp.e cuci dd, 
then will OF ed 80: 

For, applying the test of Art. 214, we have, from the first pro- 
portion, ad = be, 
and from the second, be = ad. 


RATIO AND PROPORTION. 179 


(219.) If four quantities are proportional, they will be propor- 
tional by composition ; that is, the sum of the first and second will 
have to the second the same ratio that the sum of the third and fourth 
has to the fourth. 

Let Gbsns c.2-d, 
then will a+6b:b::c+d:d. 


For, from the first proportion, i. = -. 


Add unity to each member, and we have 





a die me c+d 
that is, Ei eons cre 


(220.) If four quantities are proportional, they will be propor- 
tional by division ; that is, che difference of the first and second will 
have to the second the same ratio that the difference of the third and 
fourth has to the fourth. 


Let Ost Ga G, 
then will a—b:b::c—d:d. 


For, from the first proportion, ; = _ 


Subtract unity from each member, and we have 


c a—b c—d 
,at=z— he b rae, 7 3 


that is, a—b:b::c—d:d. 


8 








oa 


(221.) If four quantities are proportional, they will be propor- 
tional by conversion ; that is, the first will have to the difference of 
the first and second the same ratio that the third has to the cer ecence 
of the third and fourth. 


Let @ibi:erd, 
then will a:a—b::c:c—d, 


. e b d 
For, by inversion, 5: a::d:c¢; whence 7 =t 


Subtract each member from unity, and we have 








fq 
180 RATIO AND PROPORTION. 
b d a—b c—d 
1—— = 1—~— or ae 3 
a c a c 
that is, a—b:a::c—d:¢, 
or inversely, @:a—b::c:c—d. 


(222.) If four quantities are proportional, the sum of the first and 
second will have to their difference the same ratio that the sum of the 
third and fourth has to their difference. 

Let e:bsics d, 
then will at+6:a—b::c+d:c—d. 

For, by composition, a+6:6::c+d:d, 
and inverting the means,a+6:c+d::6:d. 

Also, by division, a —b:6::c—d:d, 
and inverting the means, a—b:c—d::6:d. 

Hence, by equality of ratios, a-+b:a—b::c+d:c—d. 

_(223.) If four quantities are proportional, like powers or roots 
of these quantities will also be proportional. 


Let (0408 Ceo: 
then will SSR Be 
Cage Coe eek 
For, since p= q rmasing each of these equals to the nth power, 
we have 
Gad hy 
6" 3 
that is, sb" -2 s00" a, 


where 7 may be either integral or fractional. 


(224.) If there is any number of proportional quantities all hav- 
ing the same ratio, the first will have to the second the same ratio that 
the sum of all the antecedents has to the sum of all the consequents. 

Let a, b, c,d, e, f be any number of proportional quantities, such 
that 

seQtseG 02.4: f, 


then will a@:b6::a+c+e:b4d+f. 
For, by Art. 212, we have af = be, 
ad = bc; 


also, ab = ba. 


al toy 
‘ iY ‘a 
‘ 


& 


RATIO AND PROPORTION. 181 


Therefore a(b+d+f)=b(a+c+e). 

Hence, by Art. 213,a:6::a+c+e:b+d+f. 

(225.) If three quantities are in continued proportion, the first 
will have to the third the duplicate ratio of that which it has to the 
second. 


Let etic: 0.2 Cy 
Then Gs Gs a2 0", 
For, by Art. 212, ac=0'. 


Multiplying these equals by a, we have 
. PASGh": 
1. C23 GSC x=: X Os 
Hence, by Art. 2138, a:c::a°: 8. 
(226.) If four quantities are in continued proportion, the first 


will have to the fourth the triplicate ratio of that which it has to the 
second. 


Let a, b, c,d be four quantities in continued proportion, so that 


Gia 850 42) 6 tub, 


then will TARR ML ee + 
For, by Art. 212, we have ad = de, 
ac=— 6° 3 
Also, ab = ba. 


Therefore, multiplying these equals together, we have 
a’(bdc) = b*(abc), 
or Wot OX Os 
Hence, by Art. 213, a:d::a@:08. 


(227.) If there are two sets of proportional PE the prod- 
ucts of the corresponding terms will be proportional. 


Let GAT Se CRs 

and Ooo T or Gas ibe 
Then will ae: bf:: cg: dh. 
For, by Art. 212, sad 6c, 


and ch = fa. 


182 RATIO AND PROPORTION. 
Multiplying these equals together, we have 
aexdh=—bfxcg | 
Hence, by Art. 213, ae: bf: : cg: dh. 


(228.) Three quantities are said to be in harmonical proportion 
when the first is to the third as the difference between the first and 
second is to the difference between the second and third. 


Thus, 2, 3, 6 are in harmonical proportion, for 
2:6::3—2:6—3. 

Let a, 6, c be in harmonical proportion, then | 
a:c::a—b:b—c. | 

Multiplying the extremes and means, and reducing, we have 


ab 


Gabe kas 
where c is said to be a third harmonical proportional to a and 0. 


(229.) Four quantities are said to be in harmonical proportion 
when the first is to the fourth as the difference between the first and 
second is to the difference between the third and fourth. 


Thus, 2, 3, 4, 8 are in harmonical proportion, for 
2:8::3—2:8—4 

Let a, 5, c, d be in harmonical proportion, then 
a:d::a—b:c—d. 

Multiplying the extremes and means, and reducing, we have 


ac 


Ae Qa —b’ 
where d is a fourth harmonical proportional to a, 8, c. 


(230.) Proportions are often expressed in an abridged form. 
Thus, if A and B represent two sums of money put out for one 
year at the same rate of interest, then 


A:B:: interest of A: interest of B. 


This is briefly expressed by saying that the interest varzes as 


RATIO AND PROPORTION. 183 


the principal. A peculiar character mis used to denote this re- 
lation. Thus, we write 


the interest ® the principal. 


One quantity varies directly as another, when both increase or 
diminish together in the same ratio. Thus, in the above example 
A varies directly as the interest of A. In such a case eithe 
quantity is equal to the other multiplied by some constant num 
ber. Thus, if the interest varies as the principal, then the inter- 
est equals the principal multiplied by a constant quantity, which 
is the rate of interest. 


If A ~ B, then A = mB. 


If the space (S) described by a falling body varies as the square 
of the time (T), then 


sS= mT?, 
m representing some constant quantity. 
(231.) One quantity varies znversely as another, when one in- 
creases in the same ratio that the other diminishes. Thus, the 
altitude of a triangle whose area is given, varies inversely as its 


base. 
If one quantity varies inversely as another, the product of the 


two quantities is constant. 


b 
Thus, ma X= ab. 


Conversely, if the product of two quantities is constant, then 
one varies inversely as the other. 


Thus, if AB=™m, 
m 1 
then A= B D ire 
For example, if | ry = 24, 
then ie “ak 
y 
If #=2, then y= 12, 


If == 4, théen'y— 6; 


Ae RATIO AND PROPORTION. 


my is, if x is doubled, then yis halved. The one. varies inversely 
as the other. oz 


(232.) One-quantity may vary as the prodiiprtos several others. 
Thus, in uniform motion, the space varies as the time multiplied 
into the velocity. ‘The area of a rectangle varies as the product — 
of its length and breadth. 

The weight of a stick of timber varies as its lene x its breath 
x its depth x its density. 

If the density is given, then the weight varies as the length x 
the breadth x the depth. 

If the depth also is given, then the weight varies as the length 
x the breadth. 

If the breadth is given, then the weight varies as the length. 

Finally, if the length also is given, then the weight is equal to 
a constant quantity. 


(233.) One quantity may vary directly as a second, and inversely 
asathird. Thus, according to the Newtonian law of gravitation, 
the attraction (G) of a heavenly body varies directly as the quan- 
tity of matter (Q), and inversely as the square of the distance (D). 


that ts. Ga eh 


(234.) Application of the preceding principles. 
Hes en hes A *) to find the values of x and y. 


ry — 0, 
Since ety:2::5:3. 
By Art. 220, pati. es cos 
Therefore, 3y = 2x, and y= = 


Substituting this value in the second equation, we obtain 


3" — 6, 
and x = 9. 
Therefore, e— 23, 


whence y=+2. ; 


. 


RATIO AND PROPORTION. 185 


Ex, 2. Given 2 +y:a—y::3: » (to find @ and y. 


x — y® = 56, 
From the first ‘equation, by Art. 222, we obtain 
; Qei2y:34:2; 
whence, mie: 2; 1, 
and we = Ly. 
Substituting this value of x in the second equation, we obtain 
Paes 0 4, 
Bee canons 
Ex. 3. Givenrz+y :@—y st 64: 1, } to find e-and y. 
xy = 63, 
By Art. 223, e+ y:e—y::8:1. 
By Art. 222, PEE SIATSE By ig te 
whence PN aA LS aly 
Therefore, c= ae 


Substituting this value for x in the second equation, we obtain 
Cite ome Apa geen waht B 
3 t 

: 3 Sh rev hare ait P 

Ex. 4. Given 2 — y’: x aes in 1, to find the values of x 
and y. 
Since o—y : x — 3a°y + 382y—y: : 61:1. 
3 

By Art. 220, 3xy x (r—y):e2—y ::60:1. 


2 





Hence 960: r—y :: 60: 1, 
and pentagon 
Therefore, e—y=+t 4. 
Since x? — QWvy+y’> = 16. 
Also, 4ry = 1280. 
By addition, x + Qey + y? = 1296. 
Extracting the root, w#+y=+ 36. 
Hence e= + 20, or + 16, 


y= + 16, or + 20. 


186 RATIO AND PROPORTION. 


Ex. 5. Given 2?’ —y’: a’y—ay?:: 712 
pany <6, ’{ to find and y. 
Ans. 2 =4, or 2, 
V2, or 4. 


Ex. 6. Given JV y —J/a—ra=VJVy—z2, 
GATED dade haat to find 
Vy—xe+ JSJa—x: Ja—ex::5:2, 
x and y. 


4a 5a 
Ans. ve nes base 


Ex. 7. Given2z+ J/a@:27—J2::3/2+6:2V2, to find the 
values of z. 


Ans. x =9, or 4. 


Ex. 8. What number is that to which, if 1, 5, and 13 be severally 
added, the first sum shall be to the second as the second to the 
third 2 

Ex. 9. What number is that to which, if a, b, and c be severally 
added, the first sum shall be to the second as the second to the 
third 2 

b? —ac 
Ans. a—2b+c 

Ex. 10. What two numbers are’those whose difference, sum, 

and product are as the numbers 2, 3, and 5 respectively 2 


Ex. 11. What two numbers are those whose difference, sum, 
and product are as the numbers m, n, and p? 
2p op 
Ans. ——— —. 
ns. n+m » and ae 
Ex. 12. Find two numbers, the greater of which shall be to the 
less as their sum. to 42, and as their difference to 6. 


Ex. 13. Find two numbers, the greater of which shall be to the 
less as s their sum to a, and their difference to 8. 
a+ b)? a+b 
‘A i 
ns. bye and — Tee 


Ex. 14. There are two numbers which are in the ratio of 3 to 
2; the difference of whose fourth powers is to the sum of their 
cubes as 26 to 7. Reouired the numbers. 


RATIO AND PROPORTION. 187 


Ex. 15. What two numbers are those which are in the ratio of 
m tony; the difference of whose fourth powers is to the sum of 
their cubes as p to q? 

m m? + n° n m+ n° 
Ans. — x ca li and ~2 a la 
m* —n q° m—n 


SECTION XIV. 


PROGRESSIONS. 


ARITHMETICAL PROGRESSION. 


(235.) An Arithmetical Progression is a series of quantities which 
increase or decrease by the continued addition or subtraction of the 
same quantity. 


Thus, the numbers 
113; Oy i000 Lp eee 
which are obtained by the addition of 2 to each successive term, 
form what is called an increasing Arithmetical Progression; and 
the numbers : 
20,)1'7, MAS d bso eee, 

which are obtained by the subtraction of 3 from each successive 
term, form what is called a decreasing Arithmetical Progression. 


(236.) If a represent the first term of an arithmetical progres- 
sion, and d the common difference, the successive terms of an 
increasing series will be ; 

a,a+d,a+2d, a+ 3d, a+ 4d, &c. 

The successive terms of a decreasing series will be 

a, a—d, a—2d, a— 3d, a — 4d, &e. 

Since the coefficient of d in the second term is 1, in the third 
term 2, in the fourth term 3, and so on, the mth term of the series 
will be of the form 

a+ (n— 1)d, 
which may be called the /as¢ term when the number of terms is 7. 
Hence, 


— 2 


PROGRESSIONS. 189 


The last term is equal to the first, + the product of the common dif- 
ference into the number of terms less one. 

_In what follows we shall consider the progression an ¢ncreasing 
one, since all the results which we obtain can be immediately 
applied to a decreasing series by changing the sign of d. 

If we put / to represent the last term of the series, we shall 
accordingly have 


l=a+(n— 1)d. 


This equation contains four quantities, any one of which may 
be computed when the other three are known. 


(237.) To find the sum of n cerms of a series. 
Take any series, and under it set the same terms in an inverted 
order, thus: 


Take the series } 7 ideals! “Gon ddy ld, tere 
and the same series inverted, 1g Ws Fa Gi a hs PG 
The sums are, 16, 16, 16, 16, 16, 16, 16, 16. 


The sum of the two series must be double the sum of a single 
series, and is equal to the sum of the extremes repeated as many 
times as there are terms. | 

In order to generalize this method, let S represent the sum of 
the series, 


ere a ee de a 2d od Sd ey ee +f, 
Write the same series in an inverted order, thus: 

Sal d hie Wt BS Sd ee: Ck. sat a 
Adding the two series together, term by term, we obtain 

pS Tat. (eCael Ng lg ey Aa ee tisha. 
Represent the number of terms in the series by » ; then 

28 =n(l +a). 

Hence en ree 
Therefore, 


The sum of an Arithmetical Progression is equal to half the sum 
of the two extremes, multiplied by the number of terms. 


r 


190 PROGRESSIONS. 


It also appears from the above, that the sum of the extremes is 
equal to the sum of any other two terms equally distant from the ex- 
tremes. 


(238.) The two equations 
l=a + (2 —— 1)d, 


contain five variable quantities, 

a, d, d, n, 8, 
of which any three being given, the other two may be found. 
Accordingly, 20 different cases may arise, all of which are solved 


by combining the formule above given. These cases are exhib- 
ited in the following table, and should be verified by the student : 








Required. Formule. 


i—a+(n— 1)d, 
~—=—id+ J 2dS + (a— 3d)’, 
28 
/=— —a, 
nr 
] S  (n—1)d 
2 De 


S=in} 2a-+(n—1)d, 





l+a , (l+a)(l—a 
Se ann hi aules oa 
lta 
S=—y- Xx”, 
S= in} t—(n—1)a}. 
l—a 
d=7—> 
285 — 2an 
1051) 0, 7, © = a(n — 1)’ 
d 
l 1— 
11] «2,8 jay ee 
_ 2nl— 


12) RES ic ae hy 


- 


=< _ ~ 

















PROGRESSIONS. 191 
Formule. 2a 
13} d,n, / a—l—(n— 1)d, 
S n—,1)d 
14 | d,n,8S alae 
4 dsl, S a= 30+ V (1+ 4d)? — U8, 
1o%"n, i, S 7 
nr 
ee aa 
17 | ad,l n=——+1, 
‘(2a — d)? + 8dS — d 
18 | a,d,S «Score ie AS irate 
2Qd 
28 
Tee oS Sie 
201428 Ea es ioe ve ee 12) a 
2d 
(239.) EXAMPLES. 


Ex. 1. Required the sum of 60 terms of an arithmetical pro 
gression whose first term is 5, and common difference 10. 

Ex. 2. Required the number of terms of a progression whose 
sum is 449, whose first term is 2, and common difference 3. 

Ex. 3. Required the first term of a progression whose sum is 
99, whose last term is 19, and common difference 2. 

Ex. 4. The sum of a progression is 1455, the first term 5, and 
the last term 92. What is the common difference 2 

Ex. 5. A body falls 16 feet during the first second, and in each 
succeeding second 32 feet more than in the one immediately 
preceding. If it continue falling during the space of 20 seconds, 
how many feet will it pass over in the last Becond, and how many 
in the whole time 2 f= 6 

Ex. 6. Required the sum of 101 terms of the series 


Paes, lute og mete 
Ex. 7. Find the last term of the series 


evo dy 9, 6a: 
Ans. 2n—1; 





192 PROGRESSIONS. 


that is, the last term of this series is one less than twice the number 
of terms. 
Ex. 8. Find the sum of the series ! 
1, 3, 5, 7, 9, &e. : 7 
fAns. n?; 





that is, the sum of the terms of this series 1s equal to the square of the 
number of terms. 


Thus, 1+ 3 = 42 
1+3+5 = 9=3? 
1+3+5+7 =-16—4 


14+3+51+7+4+9=25—5* 

Ex. 9. Find the sum of the natural series of numbers 
1, 2, 3, 4, 5, &e.s Ah. 

up to 2 terms. 


a. 


Ex. 10. Find the sum of the even numbers 
2, 4, 6, 8, &c., 


up to m terms. 


Ans. n(n + 1). 


Ex. 11. One hundred stones being placed on the ground ina 
straight line, at the distance of two yards from each other; how 
far will a person travel who shall bring them one by one to a 
basket which is placed two yards from the first stone? edad é : 

Ex. 12. To find m arithmetical means between two given num. “ 
bers. 

In order to solve this problem, we must first find the common 
difference. ‘The whole number of terms consists of the two ex- 
tremes, and all the intermediate terms. — If, then, m represent the 
number of means, m-+ 2 will be the whole number of terms. 

Substituting m-+ 2 for n, in formula 9, p. 222, we have 


_t—a 

m+ it 

whence the required means are easily obtained by addition. 
Ex. 13. Find 6 arithmetical means between 1 and 50. 


£5 te 


— the common difference, 


ka yg 
ae 


PROGRESSIONS. , “Nes 


Ex. 14. Find three numbers in arithmetical progression, the 
sum of whose squares shall be 1232, and the square of the mean 
greater than the product of the two extremes by 16. 

Ex. 15. Find three numbers in arithmetical progression, the 
sum of whose squares shall be a, and the square of the mean 
greater than the product of the two extremes by 0. 


Ans. 4/23" — v95,/2*; and eal 7s 


Ex. 16. Find four numbers in arithmetical progression, whose 
sum is 28, and continued product 585. 

Ex. 17. A sets out for-a certain place, and travels 1 mile the 
first day, 2 the second, 3 the third, and so on. In five days after- 
ward, B sets out, and travels 12 miles a day. How Hats will A 
creel before he is overtaken by B 2 

Ans. 8 or 15 days. 


This is another example of an equation of the second degree, 
in which the two roots are both positive. The following figure 
exhibits the daily progress of each traveler. The divisions above 
the horizontal line represent the distances traveled each day by 
A; those below the line the distances traveled by B. 


Aes ee SO TO TT 2 © 88 14 fa) 
pie ETS: Lea | | | | | he | | 
| | | | | | | | | | 

B. 1 2 3 A s 6 7 a a 10 
It is readily seen from the figure that A is in advance of B un- 

til the end of his 8th day, when B overtakes and passes him. 

After the 12th day, A gains upon B, and passes him on the 15th 

day, after which he is continually gaining upon B, and could not 

be again overtaken. 
Ex. 18. A goes 1 mile the first day, 2 the second, and so on. 

B starts a days later, and travels 4 miles per day. How long will 

A travel before he is overtaken by B2 


OaelLe W/( Shae Sab 
2 


In what case would B never overtake A? 


Ans days. 


1 


Ans. When a> ee 


N 


5 
¥ 


al PROGRESSIONS. 


For instance, in the preceding example, if B had started one 
day later, he could never have overtaken A. 


Ex. 19. A traveler set out from a certain place and went 1 
mile the first day, 3 the second, 5 the third, and so on. After 
he had been gone three days, a second sets out, and travels 12 
miles the first day, 13 the second, and so on. In how many days 
will the second overtake the first 1 


Ans. in 2 or 9 days. 


Let the student illustrate this example by a diagram like the 
preceding. 


GEOMETRICAL PROGRESSION. 


(240.) A Geometrical Progression is a series of quantities, each of 
‘which ts equal to the product of that which precedes it by a constant 
number. 

Thus, the series 


2, 4, 8, 16, 32, &c., 
and 81, 27, 9, a, Ca, 


are geometrical progressions. In the former, each number is 
derived from the preceding by multiplying it by 2, and the series 
forms an increasing geometrical progression. In the latter, each 
number is derived from the preceding by multiplying it by 4, and 
the series forms a decreasing geometrical progression. 

In each of these cases, the common multiplier is called the 
common ratio. 


(241.) To find the last term of the progression. 
Let a represent the first term, and 7 the common ratio; then 
the successive terms of the series will be 


a, ar, ar’, ar*, ar*, &e. 


The exponent of r in the second term is 1, in the ¢hird term is 
2, in the fourth term 3, and so on; hence the mth term of the 
series will be 


1 aa 


Hence, putting / for the last term, and 7 the number of terms 
of the series, we obtain 


¢=—ar'"—'. 


PROGRESSIONS. 195 


That is, 

The last term of a geometrical progression is equal to the product 
of the first term by that power of the ratio whose index is one less than 
the number of terms. 


(242.) To find the sum of all the terms. 

If we take any geometrical series, and multiply each of its 
terms by the ratio, a new series will be formed, of which every 
term except the last will have its corresponding term in the first 
series. ‘Thus, take the series 


1, 2, 4, 8, 16, 32, 
the sum of which we will represent. by S, so that 
S=142+4+4484 16+ 32. 
Multiplying each term by 2, we obtain 
28=2+4+8-+ 16+ 324 64. 


The terms of the two series are identical, except the first term 
of the first series and the /ast term of the second series. If, then, 
we subtract one of these equations from the other, all the remain- 
ing terms will disappear, and we shall have 


28—S=—64—1. 


In order to generalize this method, let a, ar, ar’, &e. , represent 
any Penterical series, and S its sum, then 


Saatar+artar-...,.. + ar? + ar"), 
Multiplying this equation by 7, we have 
rS =ar+ar+tartartt...... + ar®—' + ar". 


Subtracting the first equation from the second, we obtain 
rs —S —ar"—a. 


, a”—a 
Hence 5 = =a 
or, substituting the value of / already found, we obtain 
lr—a 
oh ey 


Hence, to find the sum of the terms of a geometrical progres- 
sion, 


196 PROGRESSIONS. 


6 * 


Multiply the last term by the ratio, subtract the first term, and di 
vide the remainder by the ratio less one. 

If a series is a decreasing one, and r consequently represents a 
fraction, it is convenient to change the signs of both numerator 
and denominator in this expression, which then becomes 





: _a—ar a—tIr 
Nnai—yp desi 
(243.) In the two equations 
bo ales 
. “r—« 
re eee Fe 


there are five variable quantities, 
a, lr, n, 8, 

of which any three being given, the other two may be found. Ac- 
cordingly, as in arithmetical progression, 20 different cases may 
arise, all of which are readily solved, with the exception of those 
in which ~ is the quantity sought. The value of m can only be 
found by the solution of an exponential equation. See Art. 336. 
These different cases are all exhibited in the following table for 
convenient reference. 














Required. Formule. 
irik S 
a+(r—1)S 
{=—_____—, etre 
T az - « 
lS — ly" — a(S—a)"**=0, © 
— (r—1Sr- ; 
rm—1 * 
a(r”? — 1 
S= ( ), 
r—l1 Ry 5 
lr-—a nt Ca 
S= 1’ eS 
S re s es hig? 
aes — oo PR 
— 1 ee 7 Seng 
{a—? pt " ‘am 
51) i a 
(7 — 1)r"-* Pes ry 
hee: aie” 
ae 
os oa 
*. » 
> 
ee 
20 
a ° Pte 
Y a 





PROGRESSIONS. 197 











Required. Formule. 
is 
= 3) | 
a 
S S—a 
r— ane o =—0; 
_S—a 
Oo ee 
rn Rs lhe tie 4, SE 
ier sey Me a cay faa 
l 
i sy 
(r— 1)S 
Tae cee as 


a = rl — (r — 1)S; 
a(S —a)’-'— US —/)*"'=0. 





_ log. i— log. a 
log. r : 
_ log. [a+ (r—1)S]—log.a 
log. r 
8 log. 1! —log. a 
7 ~ log.(S— a) —log.(S— it 1, 
_ log. l—log. [rl—(r— LS Nist 
log. 
EXAMPLES. 


Ex. 1. Required the sum of the series 
Wid, 9, 1, Ce, 
continued to 12 terms. 

Ex. 2. Required the sum of the series 
1, 2, 4, 8, 16, &c., 
continued to 14 terms, 

Ex. 3. Given the first term 2, the ratio 3, and the number pu 


terms 10, to find the last term. thant 39866 
Ex. 4. Given the first term 1, the last term 512, and she sum 
of the terms 1023, to find the ratio. = 


198 PROGRESSIONS. 


Ex. 5. Given the last term 2048, the number of terms 12, and 
the ratio 2, to find the first-term. Sy 

Ex. 6. A person being asked to dispose of his horse, said he 
would sell him on condition of receiving one cent for the first 
nail in his shoes, two cents for the second, and so on, doub- 
ling the price of every nail to 32, the number of nails in his four 
shoes. What would the horse sell for at that rate 2 

Ans. $42949672.95. 


(244.) To find any number of geometrical means between two given 
numbers. 

In order to solve this problem, it is only necessary to know the 
ratio. If m represent the number of means, m + 2 will be the 
whole number of ¢erms. Substituting m+ 2 for 2 in formula 9, 


Art. 243, we obtain 
es “Ai E 
a 


When the ratio is known, the required means are obtained by 
continued multiplication. 


~ 


Ex. 1. Find three geometrical means between 2 and 162. pe Or a”. 


Ex. 2. Find two geometrical means between 4 and 256. -“»- 


(245.) Of decreasing progressions having an infinite number of 
terms. 
The formula 
a— ar” 
sees 
which represents the sum of m terms of a decreasing series, may 
be put under the form 


In a decreasing progression, since r is a proper fraction, ”™ is 
less than unity, and the larger the number m, the smaller will be the 
quantity 7". If, therefore, we take a very large number of terms 

. . ar” ? “17 
of the series, the quantity 7”, and consequently the term par will 
be very small; and if we take 7 greater than any assignable num- 


ber, then a will be less than any assignable number. 


ys tty 


PROGRESSIONS. 199 


Hence the sum of an infinite series decreasing in geometrical 
progression is 
a 
— 
1—r 
Ex. 1. Find the sum of the infinite series 
1+$+t4+4+,&c. 


Here Gan 1, rst. 





Therefore, Ss =-—— = 


Ex. 2. Find the sum of the infinite series 
Pg eh es dee. 


Ans. 3. 

Ex. 3. Find the sum of the infinite series | 
14+itis tox t, &e. wteotelt i 
Ex. 4. Find the ratio of an infinite progression, AAS first term 


is 1, and the sum of the series &. ethiy e& 


Ex. 5. Find the first term of an infinite ph aay whose 
ratio is >, and the sum 2. ) A 


Ex. 6. Find the first paki of an infinite progression, of which 





oie 


the ratio is —, and the sum ; />- fer | ae 
n n—l oe 


(246.) PROBLEMS. 


Prob. 1. Of four numbers in geometrical progression, the sum 
of the first and second is 15, and the sum of the third and fourth 
is 60. Required the numbers. 


Let x, ry, xy’, xy’, be the numbers. 
Therefore, e+ ry = 15, 
and xy + xy*> — 60. 


eerie the first equation by 7’, 
xy’ + xy’ = 15y* = 60. 


Therefore, rom 4, 
and Uc tein ae 
Also, ete we == 1H, 


Therefore, = 5-or — 15, 


200 PROGRESSIONS. 


Taking the first value of x, and the corresponding value of y, 
we obtain the series 


5, 10, 20, 4Q; 


which numbers may be easily verified. 
Taking the second value of x, and the corresponding value of 
y, we obtain the series 


— 15, +380, —60, + 120; 


which numbers also perfectly satisfy the problem understood al- 
gebraically. If, however, it is required that the terms of the pro- 
gression be positive, the last value of x would be inapplicable to 
the problem, though satisfying the algebraic equation. 

Prob. 2.. There are three numbers in geometrical progression 
whose sum is 210, and the first exceeds 48 last by 90. What? 
are the numbers 2 i 2 @- — 

Prob. 3. There are three aah in aoe progression 
whose continued product is 64, and the sum of their cubes i is 584. 
Required the numbers. Xo 

Prob. 4. There are four numbers in areas seared progression; 
the second of which is less than the fourth by 24; and the sum 
of the extremes is to the sum of the means as 7 to 3. Required 
the numbers. 

Prob. 5. Of four numbers in geometrical progression, the. dif- 
ference between the first and second is 36, and the difference 
between the third and fourth is 4. What are the numbers? 

Prob. 6. Of four numbers in geometrical progression, the sum 
of the first and third is a, the sum of the second and fourth is 0. 
What are the numbers? 


oe . 


a® a’b ab? b° 


po ie oP et Le 


HARMONICAL PROGRESSION. 


(24'7.) 4 series of quantities is said to bein harmonical progression 
when, of any three consecutive terms, the first 2s to the third as the dif- 
ference of the first and second is to the difference of the second and 
third. 

The numbers 


60, 30, 20, 15, 12, 10, 


PROGRESSIONS. 201 


are in barmonical progression, for 


60 :.20 : : 60 — 30 :. 30 — 20 
30: 15: : 30 — 20: 20 — 15 
20: 12:: 20—15: 15— 12 
15: 10:: 15—12: 12— 10. 


So, also, the numbers 
1,2) 3) 49 $9 0) &e., 
form an harmonical progression. 


(248.) The reciprocals of a series of terms in harmonical progres ' 
ston are in arithmetical progression. 
Thus, the reciprocals of 60, 30, 20, &e., are 


being an arithmetical progression whose common difference is ;1, 


If six musical strings of equal weight and tension have their 
lengths in the ratio of the numbers 


j egestas | 


the second will sound the octave of the first ; the third will sound 
the twelfth; the fourth will sound the double octave; the fifth 
will sound the eighteenth; and the sixth will sound the third 
octave of the first. Hence the origin of the term harmonical or 
musical proportion. 

Let a, b, c be three quantities in harmonical progression, then 


a:c::a—b:b—c; 


whence | 6b = ——. 


That is, an harmonical mean between two quantities 1s equal to twice 
their product divided by their sum. 


SECTION XV. 





GREATEST COMMON DIVISOR.—PERMUTATIONS AND 
COMBINATIONS. 


(249.) THE greatest common divisor of two or more quanti- 
ties is the greatest factor which is common to each of the quan- 
tities. 


PROPOSITION. 


The greatest common divisor of two quantities is the same with the 
greatest common divisor of the least quantity, and their remainder after 
division. 

To prove this principle, let the greatest of the two quantities be 
represented by A, and the least by B. Divide A by B; let the 
entire part of the quotient be represented by Q, and the remain- 
der by R. Then, since the dividend must be equal to the prod- 
uct of the divisor by the quotient + the remainder 


Now every number which will divide B will divide QB; and 
every number which will divide R and QB will divide R + QB or 
A. That is, every number which is a common divisor of B and 
R is a common divisor of A and B. 

Again, every number which will divide A and B will divide A 
and QB; it will also divide A— QB or R. That is, every num- 
ber which is a common divisor of A and B is also a common 
divisor of B and R. Hence the greatest common divisor of A 
and B must be the same as the greatest common divisor of B 


and R. 


(250.) To find, then, the greatest common divisor of two quan- 


GREATEST COMMON DIVISOR. 203 


tities, we divide the greater by the less ; and the remainder, which 
is necessarily less than either of the given quantities, is by the 
last article divisible by the greatest common divisor. 

Dividing the preceding divisor by the last remainder, a still 
smaller remainder will be found, which is divisible by the greatest 
common divisor; and by continuing this process with each re- 
mainder and the preceding divisor, quantities smaller and smaller 
are found, which are all divisible by the greatest common divisor, 
until at length the greatest common divisor must be obtained. 
Hence the following 


RULE. 


Divide the greater quantity by the less, and the preceding divisor by 
the last remainder, till nothing remains ; the last divisor will be the 
_greatest common measure. 

When the remainders decrease to unity, the given quantities 
have no common divisor, and are said to be imcommensurable, or 
prime to each other. 


EXAMPLES. 


Ex. 1. What is the greatest common divisor of 372 and 246 2 
372 | 24.6 


246) 1 
24.6 | 126, first Rem. 
204 ie ae 
126 | 120, second Rem. 
120) 
120 |__6,third Rem. 


120 | 20 





Here we have continued the operation of division until we ob- 
tain 0 for a remainder ; the last divisor (6) is the greatest common 
divisor. Thus, 246 and 372 being each divided by 6, give 41 and 
62, and these quotients are prime with respect to each other; that 
is, have no common divisor greater than unity. 


Ex. 2. What is the greatest common divisor of 
336 and 7201 

Ex. 3. What is the greatest common divisor of 
918 and 5221 


204 GREATEST COMMON DIVISOR. 


(251.) In the application of this rule to polynomials, some 
modification may become necessary. It may happen that the 
first term of the dividend is not divisible by the first term of the 
divisor. This may arise from the presence of a factor in the 
divisor which is not found in the dividend, and may therefore be 
suppressed. For, since the greatest common divisor of two quan- 
tities is only the product of their common factors, it cannot be 
affected by a factor of the one quantity which is not found in the 
other. f 

We therefore suppress in the first polynomial all the factors 
common to each of its terms. We do the same with the second 
polynomial, and if the suppressed factors have a common divisor, 
we reserve it as forming part of the common divisor sought. 

But if, after this reduction, the first term of the dividend, when 
arranged according to the powers of some letter, is not divisible 
by the first term of the arranged divisor, it follows from the above 
that we may multiply the dividend by any monomial factor which 
will render its first term divisible by the first term of the divisor. 


EXAMPLES. 


Ex. 1. Required the greatest common divisor of 
a+ x* and «*t— 1. 
2 + a | at*— 1 
x—x |x 
WALID 5 +2, first Rem 
Suppressing 2, we have a?+1. 





at — 1| x + 1 
a+ x |2’?—1 
—r’—1 
—a’— 1, 


Whence 2*+ 1is the greatest common divisor. To verify this 
result, divide x° + 2° by 2’? -+ 1, and we obtain 2°; divide at—1 
by x’ + 1, and we obtain x2? — 1. 

Ex. 2. Required the greatest common divisor of 

x* — bx and a? + 2dr + 0B’. 
xo — bax | 2° + 2ba + 8 
x + 2bx* + Ba oe ets 
— 2x’ — 2b’, first Rem. 





PERMUTATIONS AND COMBINATIONS. 205. 


Suppressing the factor — 2dz, we have « + 8. 
vt er +0 |x+b 
a? + bx e+tob 
be + & 
bx + b*. 
Whence z + 6 is the greatest common divisor. 
Ex. 3. Required the greatest common divisor of 


4a° — 2a? — 3a+.1 and 3a° — 2a — 1. 
Ans.a—1. 





Ex. 4. Find the greatest common divisor of 
a2? —1andaxr+1. 
Ex. 5. Find the greatest common divisor of 
x* — a‘ and x — a’. 
Ex. 6. Find the greatest common divisor of 
a? — 3ab + 20? and a? — ab — 206°. 
Ex. 7. Find the greatest common divisor of 
a‘ — x* and a — ax — az’ + o*. 
Ex. 8. Find the greatest common divisor of 
a’ — ab + 3ab? — 33° and a? — 5ab+ 40". 


THEORY OF PERMUTATIONS AND COMBINATIONS. 


(252.) The different orders in which quantities may be arranged 
are called their Permutations. Thus, 

a, b, ¢, 

\: G0, 

the permutations of the three letters a, b, c, taken all to- j 5, a, c, 

gether, are ‘ : : : : nS ; bake’ Oa Get, 

| pats aod 

(ty bya. 


( a, b, 

a, C, 

The permutations of the same letters taken two and two, J b, a, 
are . . . : . ° . . . . 14,6, 
C, a, 

c, b. 


206° PERMUTATIONS AND COMBINATIONS. 
The permutations of the same letters taken singly, or one\ 
by one, are ; : : ; : . , 
Cs 


(253.) To find the number of permutations of n letters, taken m and 
m together. 

Let a, b,c,d.....k, be the n letters. 

The number of permutations of m letters taken singly, or one 
by one, is evidently equal to the number of letters, or to n. 

The number of permutations of m letters taken two and two is 
n(n — 1). For if we reserve one of the letters, as a, there will 
remain 72 — 1 letters. 


Dy 0, Oe hes 
Writing @ before each of these letters, we shall have 
Q0,9A0, 100 se wisest 


that is, we obtain 7 — 1 permutations of the m letters taken two 
and two, in which a stands first. Proceeding in the same manner 
with b, we shall find 2 — 1 permutations of the z letters taken two 
and two, in which 0 stands first ; and so for each of the x letters. 
Hence the whole number of permutations will be 


n(n — 1). 


The number of permutations of m letters taken three and three 
together is 


n(n — 1) (n — 2). 


For if we reserve one of the letters, as a, there will remain 
m—1 letters. Now we have found the number of permutations 
of x letters taken two and two to be m(n— 1). Hence the per- 
mutations of » -—-1 letters taken two and two must be 


(27 — 1) (n— 2). 


Writing a before each of these permutations, we shall have 
(x — 1) (x — 2) permutations of the m letters taken three and 
three, in which a stands first. Proceeding in the same manner 
with b, we shall find (n— 1) (x — 2) permutations of the z letters 
taken three and three, in which 0 stands first ; and so for each of 
the letters. Hence the whole number of permutations will be 


n(n — 1) (n— 2). 





bs 


PERMUTATIONS AND COMBINATIONS. 907 


In like manner, we can prove that the number of permutations 
of m letters taken four and four is 


n(n — 1) (n— 2) (n— 3). 


When the letters are taken two and two, the last factor in the 
corresponding formula is »—1. When the letters are taken 
three and three, the last factor is » —2. When the letters are. 
taken four and four, the last factor is 2 — 3. 

Hence, when the letters are taken m and m together, the last 
factor will be 2 —(m— 1) orn—-m-+1; and the number of per- 
mutations of z letters taken m and m together will be 


n(n— 1) (n—2) (n—3)..... (7 — m+ 1). 
EXAMPLES. | 


Ex. 1. Required the number of permutations of the 8 letters 
a, b, c, d, e, f, g, A, taken 5 and 5 together. 
Here n=8,m=5,n—m+1=4, 
and the above formula becomes 
8.7.6.5.4 = 6720, Ans. 


Ex. 2, Required the number of permutations of the 26 letters 
of the alphabet, taken 4 and 4 together. a 


Ex. 3. Required the number of permutations of 12 letiers, rer 
6 and 6 together. Ag. 


ae So 


(254.) If we suppose that each permutation comprehends all the n 
letters ; that is, if m =n, the preceding formula becomes 


n(n —1)(m—2)..... 2x1; 
or, inverting the order of the factors, 
1 DE Pe. (n —1)n ; 


which expresses the number of permutations of 7 letters taken 
all together. 

Ex. 1. Required the number of changes which can be rung 
upon 8 bells. 

According to the preceding formula, we have 


1.2.3.4.5.6.7.8 = 40320, Ans, 


Ex. 2, What is the number of permutations which jag be 
formed from the letters composing the word “ virtue ?” y 


» 


208 PERMUTATIONS AND COMBINATIONS. rs, 
ek: 


Ex. 3. What is the number of different arrangetestl which 


can be made of 12 persons at a dinner-table 2 37%, Ze) be 


~ (255.) The combinations of any number of quantities signify the 
different colleétions which may be formed of these qiantitiee, 
without regard to the order of their arrangement. 

Thus, the three letters a, b,c, taken all together, form but one 
combination, abc. 

Taken two and two, they form three combinations, 


ab, ac, be. 


To find the number of combinations of n letters, taken m and m to- 
gether. 

The number of combinations of nm letters taken separately, or 
one by one, is evidently n. 

The number of combinations of x letters taken two and two, is 
n(n — 1) 

Vi 

For the number of permutations of 7 letters taken two and two 
is n(m — 1); and there are two permutations (ad, ba) correspond- 
ing to one combination of two letters. Therefore the number of 
combinations will be found by dividing the number of permuta- 
tions by 2. 3 

The number of combinations of m letters taken three and three 
n(n — 1) (2 — 2) 

1.2.3 ; 

For the number of permutations of 7 letters taken three and 
three, is »(z — 1) (2 —2); and there are 1.2.3 permutations for 
one combination of three letters. Therefore the number of com- 
binations will be found by dividing the number of permutations 
by 1.2.3. 

In the same manner, we find the number of combinations of n 
letters, taken m and m together, 


n(a—1) (mn —2).....(m—m-+ 1) 
Ex. 1. Required the number of combinations of six letters taken 
three and three together. 
Here n=6, m=3,n—m+1=4, 
and the formula becomes 





together, is 





iy 


~ Me 4 rs we 
* ip 


PERMUTATIONS AND COMBINATIONS. 209 


Ex. 2. Required the number of combinations of 8 letters taken 

- dand 4. : | Loe. fil 

ee Fist 3. Required the number of combinations of 10 letters taken 
6 and 6. hr tie Qi 0 

The following table, which is computed by the preceding for- 

mula, shows the number of combinations of 1, 2, 3, 4, &c., letters 
taken singly, or two and two, three and three, &c. An important . 

application of this table will be seen in the next Section. 


5 and 5.|6and 6. 


Number of combinations. 










Lett’rs.| Singly.}2 and 2.|3 and 3.|4 and 4. 






7and7. 





8 and 8.|9 and 9./10, 10. 
































SCOMDNIAUMPEP WHR 
COMVIOURWWDe 


a 
—_ 


SECTION XVI. 


INVOLUTION OF BINOMIALS. 


(256.) We have shown, in Art. 142, how to obtain any power 
of a binomial by actual multiplication. We now propose to de- 
velop a theorem by which this labor may be greatly abridged. 

Taking the binomial a+ 4, its successive powers found by 
actual multiplication are as follows: 


(a + b)} = @ + b, 

(a+ bP =a’ + 2ab + 2B, 

(a+ bf =a’ + 307) + 3a? + 8%, 

(a+ b)*=a*+ 40°) 4+ 607? +400? + 04, 

(a + 6)’ =a* + 5a‘d + 100°D? + 10078? + 5ab* +8, 

(a+ 6)’ =a° + 60° + 15a*b? + 200°b? + 1507b* + Gab? + B°. 


(257.) The powers of a — 0, found in the same manner, are as 
follows : 


(a—b)'=—a —), 

(a— bP =a — 2ab +8’, 

(a— bf’ =a — 3a°) 4+ 300? ~—D*, 

(a— b)* =a*— 40°) + 60°? —4ab? + 0%, 

(a — b)’ =a’ — 5a‘d + 100°? — 10a°b° + 5ad* ~— 0’, 

(a — b)’ = a® — 6a°d + 15a*d? — 20a°b? + 15a7b* — Gab? + b°. 


On comparing the powers of a+) with those of a—8, we per- 
ceive that they only differ in the signs of certain terms. In the 
powers of a+ J, all the terms are positive. In the powers of 
a—b, the terms containing the odd powers of d have the sign —, 
while the even powers retain the sign +. The reason of this is 
obvious; for since — 0 is the only negative term of the root, the 


INVOLUTION OF BINOMIALS. 211 


terms of the power can only be rendered negative by 6. A term 
which contains the factor —} an even number of times, will there- 
fore be positive; if it contain it an odd number of times, it must 
be negative. Hence it appears that it is only necessary to seek 
for a method of obtaining the powers of a+ 0; for these will 
become the powers of a — 3, by simply changing the signs of the 
alternate terms. 


(258.) If we consider the exponents of the preceding powers, 
we shall find that they follow a very simple law. Thus, 


9 
In the square, the exponents . . \ y ; wa 0, 4 
In the cube, the exponents i ofaare 3, 2, 1, 0, 
of, dbiare  °0)"1)'2," 3: 
In the fourth power, tne exponents | ofaare 4, 3, 2, 1, 0, 
of, Dare .0y, J 2s ede 
&c., Sc. &e 


In the first term of each power, a is raised to the required 
power of the binomial ; and in the following terms the exponents 
of a continually decrease by unity to 0; while the exponents of 
6 increase. by unity from 0 up to the required power of the bino- 
mial. The sum of the exponents of a and 0 in any term is equal 
to the exponent of the power required. ‘Thus, in the second 
power, the sum of the exponents of a and d in each term is 2; in 
the third power, it is 3; in the fourth power, 4, &c. 

We hence infer that for the seventh power the terms, without 
the coefficients, are 

a’, a°b, a°b’, a*b®, a°b*, ab’, ab®, 0" ; 
and for the mth power 
ee eo Oe mr gna 4 Glas aba *® Ds. 


(259.) It remains to determine the coefficients which belong to 
these terms ; and in order to discover the law of their formation, 
let us take the coefficients by themselves. 


The coefficients of the 1st power are bony 
The coefficients of the 2d power are Lie 2m 41 
The coefficients of the 3d power are HGNDUCZ00 9 


The coefficients of the 4th power are 1. 4°. 6. 14 ed 
The coefficients ofthe 5th powerare 1 5 10 10 5 1 
The coefficients ofthe 6th powerare 1 6 15 20 15 6 1 


212 INVOLUTION OF BINOMIALS. 


The numbers in this table are identical with those in the table 
of combinations on page 209. For example, the coefficients of 
the fifth power denote the number of combinations of five letters 
taken one and one, two and two, &c.; the coefficients of the sixth 
power denote the number of combinations of six letters taken one 
and one, two and two, &c. The reason of this will appear if we 
observe the law of the product of several binomial factors, © + a, 
x+b,x+c, x+-d, &e. 


Multiply «+a 

by z+ b, 

we obtain a -+(a-+b)a-+ab=I1st product. 
Multiply by x + ec, 

we obtain “e+(a+b +c)x*-+ (ab+-ac-+ bc)x+ abc = 2d product. 
Multiply by x + d, | 

we obtain at+(a+b+c+d)x°+ (ab+ac+ad-+ be-+ bd-+ cd)x?+ 

+ (abe+ abd+ acd-+-bcd)x-+ abcd =3d product. 


We observe that in each of these products the coefficient of x in 
the first term is unity ; the coefficient of the second term is the sum of 
the second terms of the binomial factors ; the coefficient of the third 
term is the sum of all their products two and two ; the coefficient of the 
fourth term is the sum of all their products three and three, &e. 

It is easily seen that if we multiply the last product by a new 
factor, x + e, the samie law of the coefficients will be preserved. 

If now, in the preceding binomial factors, we suppose a, J, ¢, d, 
&c., to be all equal to each other, the product 


(2+ a) (e+6) (e+ c)(e«+d)..... 
becomes (2 +a)”. 

The coefficient of the second term of the product, ora+b+c+d 
Hs 46 , becomes a+a+a-+a.....; that is,a takenas many times 
as there are letters a, b, c, d, and is consequently equal to na. ~ 

The coefficient of the third term, or ab + ac, &c., reduces to 





@ta+a’..... , or a’ repeated as many times as there are dif- 
ferent combinations of 7 letters taken two and two; that is, 
able wel) Gr 

1.2 . 


The coefficient of the fourth term reduces to a* repeated as 


INVOLUTION OF BINOMIALS. 213 


many times as there are different combinations of v letters taken 
n(n — 1) (n — 2) 
1.2.3 
Thus, we find that the mth power of x + a may be expressed as 
follows : 


three and three, that is, a’, and so on. 


—1) 2 2 n(n— 1) (n —2) wea" 

ROM oH on Haas rf 

+na"'x+a", 
which is called the Brnom1at THeoreEm, and is generally ascribed 
to Sir Isaac Newton. So important was it regarded, that it was 
engraved on his monument in Westminster Abbey as one of his 
greatest discoveries. 

On comparing the different terms of this development, we per- 
ceive that any coefficient may be derived from the preceding one 
by the following rule: If the coefficient of any term be multiplied by 
the index of x in that term, and divided by the index of a increased by 
one, it will give the coefficient of the succeeding term. 

Thus, the fifth power of x + a is 

x + 5ax* + 10a°x* + 10a°x* + 5a*x + a’. 

If the coefficient 5 of the second term be multiplied by 4, the 
index of x in that term, and divided by 2, which is the index of a 
increased by one, we obtain 10, the coefficient of the third term. 

So, also, if 10, the coefficient of the fourth term, be multiplied 
by 2, the index of x, and divided by 4, the index of a increased 
by one, we obtain 5, the coefficient of the fifth term; and so of 
the others. 

The coefficients of the sixth power will also be found as fol- 
lows: 


(v+a)"=2"+nax"—!-+4+ ——_~_— sa olive 


1, 6, 9X5, 1x4, 20x38 15x2 6x1. 
ip Shp: 4) 5 6? 
that is, Peete. Ot a Pos autora Te 
The coefficients of the seventh power will be 


7X6 215 35x4 35x3 21x2 7x1 


DW eee ee? > Dla Tee > ae 7 3 
1 


1 Ay 








ea: 4 Bui 1 6 

(hawied Weve, Biv, eabe ut 635, 0 BEL AH, 
Therefore, the seventh power of 7 + a is 

x’ + Tax’ + 21a’a® + 35a%x'* + 35a1x® + Qlara? + Ta°e + a’. 


, awe Ss oe ‘ 
4 ey Te 
; . 
‘ 


214, INVOLUTION OF BINOMIALS. 


It is sometimes preferable to retain the factors of the coeffi- 
cients distinct from each oie as follows: 


76° 168.) 16.64. eR 4S 
1__ pl 6 yj SRY ae oat tt fs 2 
(2-14) ao pia +o gt +79 3°" +79 347% ay hee 
7.6.5.4.3.2 5 | 7.6.54:3.201 


19.34.56°"+1934367"" 


The factor 1 is retained for the sake of symmetry, and to ex- 
hibit more clearly the law of the coefficients. 


(260.) The theorem thus developed is expressed in the following 


RULE. 


In any power of a binomial x + a, the index of x begins in the first 
term with the index of the power, and in the following terms continu- 
ally decreases by one. The index of a commences with one in the 
second term of the power, and continually increases by one. 

The coefficient of the first term is one ; that of the second is the 
index of the power ; and if the coefficient of any term be multiplied by 
the index of x in that term, and divided by the index of a increased by 
one, it will give the coefficient of the succeeding term. 


(261.) The nawmber of terms in the power is always greater by 
unity than the index of the power. ‘Thus, the number of terms 
in (a+ b)* is 4+ 1 or 53 in (a+ 6)° is 6 + 1 or 7. 

Also, if we examine the table in Art. 259, it will be perceived 
that after we pass the middle term, the same coefficients are re- 
peated in the inverse order. Thus, the coefficients of 

(a+ 5)’ are 1,90, OMI Ofer: 
of (a+ 6)° are Tp OLD Y 220 10.0 a. 

Hence it is only necessary to compute the coefficients for half 
the terms; we then repeat the same numbers in the inverse order. 

(262.) The sum of the coefficients for each power is equal to the 
number 2 raised to the same power. For, let r= 1 anda=1, 
then each term without the coefficients reduces to unity, and the 
value of the power is simply the sum of the coefficients. Also, in 
this case, (c+ a)" becomes 2”. Thus, the coefficients of the 


first power are 1-1-2 2, 

second power are 14+2+4+1=4=2%’, 

third power are 1+4+3+4+3+4+1=8=2%, 

fourth power are 14+4+64+441=16=2%. 
&c., &ec., SG 


‘4 


INVOLUTION OF BINOMIALS. 215 


EXAMPLES. 


Ex. 1. Raise 2 + a to the 9th power. 
The terms without the coefficients are 


9 8 2 aT 3 6 4 5 ant 6 3 T m2 8 9 
mt, an’, aa", aa, a’, ara, aba, aa", ae, a’. 
And the coefficients are 


9x8 36x7 84x6 1265 126x4 84x3 36x29x1 
iL we ear ane Fic 7 ka Tare y OO ee ae ee Ieee are beeeea ert pr; oe ae ae. 
2 3 4 
Sint IB 1, 9.) 86) bu B4y6 AI2BY em 126yii84yr 20 B6j90 15 OY 01 11: 
Prefixing the coefficients, we obtain 
(x+a)°=2°+ 9ax*+ 36072" + 84a7x°+ 126a*x°+ 1260°x*+ 840%? + 
+ 36a'x’+ 9a®x+a’. 
It should be remembered that, according to Art. 261, it is only 
necessary to compute the coefficients of half the terms independ- 
ently. 


Ix. 2. What is the 6th power of r —a? 


(263.) If the terms of the given binomial are affected with 
coefficients or exponents, they must be raised to the required 
powers, according to the principles already established for the 
involution of monomials. 

Ex. 3. Raise 2x + 5a’ to the fourth power. 

For convenience, let us substitute } for 2x, and c for 5a’. 


Then (b+ c)*=b*+ 4b%c + 6B'c? + 4bc? + ct. 
Restoring the values of 6 and c, 
The first term will be (2x)* == LOds 


The second term will be 4(2z)?. 5a? —4.8.52°a’. 

The third term will be 6(2a)?.(5a°)?=6.4.2527a'. 

The fourth term will be 4(2x) . (5a’)’=4.2.125za". 

The fifth term will be OL UOd ee Onn: 

Therefore, (2a-++5a’)*= 16x*+ 1602°a’+ 600x°a*-+ 1000xa°+ 625a°. 
Ex. 4. What is the fourth power of 2x°+ 5y’? 

Ex. 5. What is the seventh power of 2a + 36? 

Ex. 6. What is the sixth power of a’ + 3ab? 

Ex. 7. What is the fourth power of 5c’ — 2yz? 


(264.) By means of the Binomial Theorem we can raise any 
polynomial to any power. 





216 INVOLUTION OF BINOMIALS. . 


For example, let it be required to raise a+ b+ c¢ to the third 


power. 4 
For convenience, we put b+c—=m; we then obtain 


(a+b+cyP=(a+m) =a’ + 8a'm + 38am’ + m’. 
Substituting for m its equal, b + c, we obtain 
(a+b+c) =a’ + 3a°(b+ c) + 3a(b+ cc)? + (b+ cy. 
It only remains to develop the powers of the binomial 6 + c, 


and to perform the multiplications which are indicated. We thus 
obtain 


SS ae 


(a+b+ c)y=a’ + 3a°b + 3ab ’ + B, 
+ 38a’c + 6abc + 306’c, i 
-+38ac? + 3c’, ' 
+c, | =e 
Ex. 2. Raise x + a+ 6b to the fifth power. XH FF Eo" +) O20}, 
(265.) When one of the terms of a binomial is unity, the powers 
assume a simpler form, since every power of 1 is 1. 
Thus, the fourth power of a + 0, which is 
at + 4.a°b + 607b? + 4ab* + dF, 
when we make a = 1, becomes 


1+ 40 + 60° + 40° + dt. 


ee ee ee ee 


So, also, (1-++a)” =1+4 a+ Wes aie a’ Bera ieee 


Every binomial of Be form v3 + a)" may be reduced to the 
form of 2"(1+ 0)". For 


eta =@ (4+$). 


Therefore, (x-+-a)"=2” (1+ =i or, putting d= * 


=2"(1+6)", 
=a'[14n.2 amt uae n(n— 1) (n—2) a? 











pans pot tke], 

This expression for the value of (x + a)” is equivalent to that 
on page 213, as may be easily shown by multiplying 2” into each 
term within the parenthesis. For some purposes this is regarded 
as the simplest form. 


(266.) In the development of the binomial (vx+a)", we have 





— 


INVOLUTION OF BINOMIALS. 217 


hitherto supposed 7 to be a positive integer. The Binomial Theo. 
rem is, however, applicable, whatever be the nature of the quan- 
tity m, whether it be positive or negative, integral or fractional. 
When 7 is a positive integer, the series consists of 2+ 1 terms. 
In every other case, the series never terminates ; that is, the de- 
velopment of (x + a)" furnishes an infinite series. 


. : 1 
Ex. 1. It is required to convert ———— or (a+ 6)~’ into an in- 


a+ 6 


finite series. 
According to Art. 260, the terms without the coefficients are 
a fata, Gran, alt, a—°b', an’, &e. 
The coefficient of the first term is 1. 
The coefficient of the 2d term is — 1, the index of the power. 


The coefficient of the third term is aE asneeeitics +1. 

The coefficient of the fourth term is 7} _ Peis 
f aor —l1x—4 

The coefficient of the fifth term is Wee aoa hc 41. 

The coefficient of the sixth term is it - Aa ——1. 


We thus obtain 

1 
a+b 
where the law of the series is obvious; the coefficients are all 
unity, and the signs are alternately positive and negative. 

We might have obtained the same result by the ordinary metho? 
of division. The operation is as follows: 


=(a+ 6) =a" '— ab 4 a * 0? — a4? ++- a — a 0° 4, &e., 








1 a+b 
b 1s & ; 
1+- a gt gt gi + en = the quotient. 
b 
Waa 
ye 
a @ 
5? 
+B 
Pia 
He oid oa 
53 


218 INVOLUTION OF BINOMIALS. 





Hence, 
1 Ye yi eee " tee b 4 
——_ = -— = + =— = &e.; which may b itten | 
atb~ a php pee gti” Mp | “2% 
a-*— a~°*b + a “bh? — a~*h? +, &e., the same as before found; ‘a 
and it is obvious, from inspecting the operation . division, that 
the series will never terminate. pil = 4 ; ¥ ‘ 
1 : : - 
Ex. ital ired to convert ——.. or (a+b)? into anin-= 
x is require conver (a-Lb) (a+ ) into ani | d 
finite series | ner i fae ow a 
1°) 26 2630 4b" Bp Se Nes ke 
Ans. — —S tas Leer Pe FOES 
a’ a a’ 6 ie 
or, @ *°— 2a *b + 3a7*b’? — 4a~°b* + 5a °O, te it + 


Here the coefficients increase regularly by 1, and the sions are a ng 
alternately positive and negative. We might have obtained the 
same result by division, as in the former example __ ‘ 





1 
Ex. 3. Expand into a series rae Oe ieee) bg gebee me 


. 


Ex. 4. Expand into a series —— Ce. —; or (a—b)-*%, 99 ob ee st 4 
; Q 4 } - 4 fee ae ab 
Ex. 5. Expand into a series (a + oe 0 a MOR See rian s 
2 5 : ~~ 3 4/7) 6/1 4 L 
Ex. 6. Expand into a series (a—b)-*, 2 734% Oe se) i 
(267.) We have now considered the powers of a binomial when ie 
the exponent is an integer, either positive or negative. It re- As 3 
mains to consider the case when the exponent is a fraction. - ber 
EXAMPLES. 


ee iL 
Ex. 1. Expand / a+ or (a+ 6)? into an infinite series. 
The terms without the coefficients are 
1 
1 @,a 4b, a 25° , a eb: a — 354, &e. 


The exponents of a decrease by unity, while those of 6 in- 
crease by unity. 


The coefficient of the first term is i | 
The coefficient of the second term is +4. 

: ia 
The coefficient of the third term is Se 





INVOLUTION OF BINOMIALS. 219 





a ia Sis tiede i 1.3 
The coefficient of the fourth term is 5 —-+ 9.4.6" 
| 1.3 
| 5 aiihietsk ao gla ede 
The coefficient of the fifth term is z =~ 548 
The series then is x 
(a+ Be her te Ne ga ah, 13.9 2h! +; &e. 


2 24 2.4:.6 2.4.6.8" 


The factors which form the coefficients are kept distinct, in 
order to show more clearly the Jaw of the series. The factors 
in the numerator compose the series of odd numbers, 1, 3, 5, 7, 
&e. ; while the denominator contains the even numbers, 2, 4, 6, 8, 
10, &c. 

The above series expresses the square root of a+ 6. We 


should obtain the same result if we extracted the square root by 


the usual method. See Art. 286. 
Ex. 2. It is required to convert (a? + «)’ into an infinite series. 


N Pim A a 32° 3.52% ‘ __ 3.5.70" Res 
; ns. a+ nie res OA ka) 24.08.10a oh 
ax. ocx Sax 3.5.a7 "4 3.5.7a 2 
or, Setco og om 24.6 2.4.6.8 9.4.6.8.10 —° &e., 


where the law of the series is evident. 
Ex. 3. It is required to convert (a— 2)? into an infinite series. 
Ex. 4. It is Paty to convert (a + b): into an infinite series. 
26? 2.55% 2.5.85 
+, &c. 2 


1 eye Freee 
Ans. a aa, : 3.60" 3.6.90 3.6.9.12a" 


Ex. 5. Expand (a — 6)* into an infinite series. 

2 3 4 
Ans. a* ; 1 — se — ae —< ek 3 pane ; —, &e. ; : 
Ex. 6. Expand (a + 2)? into an infinite series. 2 1 ioe Hi: 
Ex. 7. Expand (1— 2)5 into an infinite series. /- ~ 
Ex. 8. Expand (a? + 8°)! into an infinite series. 
Ex. 9. Expand (a? — 8°) into an infinite series. 
(268.) The binomial theorem is applicable to cases in which 

the value of the exponent v is a negative fraction. 





220 | INVOLUTION OF BINOMIALS. SS 


EXAMPLES. 


< ~] 
. Seti centeet —4. 
Ex. 1. Expand into a series (a+ 5) or (a+) 
The terms without the coefficients are 


1 3 


—_— AL 


2a bh ab a 1), a 3's ae 
Der aie De ye} , a Nee , &e. 


The coefficient of the first term is i| 
The coefficient of the second termis —1. 
1 yee 
The coefficient of the third term is : , 2—+ = 
1:3 
— X—s 
Th ficient of the fourth t a Joins SRD 
e coefficient of the fourth term is 5 = OTe 
1.3.5 
[246% — 3 | eae 
The coefficient of the fifth term is 4 =+55G8° 


Hence we obtain 





—, &e. 


Ex. 2. Expand into an infinite series (¢ + x2)! 


ra Nias We Sige BE a a 14 ts 4.7.10 —13 
Ans. a > *—-a3@ ay Py eee 


3 BiG te. bh 1au6 Ak ait a aiGra uke , 
&e. 


Ex. 3. Expand (1+ x) into an infinite series. 


2 3 oe 
Ans. 1—2 4 6% _ S11? | 6.11.162 


Bt 5.0 5.1018 7 601690) oe 


Ex. 4. Expand (a?— 2)? into an infinite series. 


—_——~—— jnto'an infinite series. 
Jb +c 


Ex. 5. Expand 





(269.) The binomial theorem may be employed to determine 
the roots of surd numbers. 


EXAMPLES. 


Ex. 1. It is required to find the square root of 2. 


INVOLUTION OF BINOMIALS. opt 


- The development of (a + b)' has been given in Example 1, page 
219. If we make a = 1 and b=1, then (a + b)* becomes (1+ 1)! 
or / 2; and the terms of the development become 


1 1.3 aks) 1.3.5.7 


1 
1+35—94+ a467 24.68 + 24.68.1077 © 


which therefore expresses the square root of 2. The sum of this 
series is 1.41421. As, however, the series does not converge 
rapidly, it would require a large number of terms to give the root 
with tolerable accuracy. The following example affords a better 
illustration of the wézlity of the method. 


Ex. 2. Required the square root of 101. 


101 = 100 (14+ ma) Therefore / 101 = 10 (+ a) 


ated ==) 1 and.6 = aes in the development of (a + 6)' on page 


251, and we shall have 


1 1.3 1.3.5 
v101=10 (45 100 24.100°' 2.4.6.100° 2.4.6.8.100°'? &c. 


This series converges so rapidly that the first two terms givea 
result correct to three decimal places ; and five terms give a re- 
sult correct to ten decimal places. 


Thus, the value of the first term is 1.00000000000 


The value of the second term is + .00500000000 
The value of the third term is — .00001250000 
The value of the fourth term is + .00000006250 
The value of the fifth term is . — .00000000039 

Their sum is * "1,00498756211 


And multiplying by 10, we have 
J 101 = 10.0498756211. 


Ex. 3. It is required to convert *\/9, or its equal (8 + 1)', into 
an infinite series, and find its value. 


1 5 5.8 5.8.11 
Ans. 2tap 3.6.2 36.9.9' 3.6.9.12.29' 36.9.19.15.28 
— 9.08008. 


y OCCas 






sn 












bh ae i a ay a 
ct 
ad 
‘ , ; ol : 
“ 222 -INVOLUTION OF BINOMIALS. 
Ex. 4. It is required to extract the cube root of 31. | 
a \ | 
Bix" VOT FE =v 27 (1+ ny 
4 2.47 2.5.4° 2.5.8.4¢ 
; = It 327 TSERF IIT 6ILFS SEM IRR Ts si 
= 3.14138, 
n 
; ( 
; 


SECTION XVII. 


EVOLUTION OF POLYNOMIALS. 


(270.) Method of extracting the square root. 

In order to discover a rule for extracting the square root, let 
us consider the square of a+ 0, which is a+ 2ab+ 0. If we 
write the terms of the square in such a manner that the powers 
of one of the letters, as a, may go on continually decreasing, the 
first term will be the square of the first term of the root; and 
since in the present case the first term of the square is a’, the 
first term of the root must be a. 

Having found the first term of the root, we must consider the 
rest of the square, namely, 2a) + 4’, to see how we can derive 
from it the second term of the root. Now this remainder, 2ab-+-2”, 
may be put under the form (2a-+ 6)b; whence it appears that 
we shall find the second term of the root if we divide the re- 
mainder by 20+. The first part of this divisor, 2a, is double of 
the first term already determined ; the second part, 6, is yet un- 
known, and it is necessary at present to leave its place empty. 
Nevertheléss, we may commence the division, employing only the 
term 2a; but as soon as the quotient is found, which, in the pres- 
ent case, is 6, we must put it in the vacant place, and thus render 
the divisor complete. 


The whole process, therefore, may be represented as fol- 
lows: 


a+ abt & ja+b= the root. 
a 


Qab + b | 2a-+6=the divisor. 
2ab + & | 


224 EVOLUTION OF POLYNOMIALS. 


Hence we derive the following 


RULE FOR EXTRACTING THE SQUARE ROOT OF A 
~ POLYNOMIAL. 


Arrange the polynomial according to the powers of some one Letter ; 
take the root of the first term for the first term of the fe root, and 
subtract its square from the given quantity. 

Divide the first term of the remainder by double the root already 
found, and annex the result both to the root and the divisor. Multiply 
the divisor thus increased by the last term of the root, and subtract the 
product from the last remainder. Proceed in the same manner to find 
the additional terms of the root. 

Ex. 1. Required the square root of a* — 2a°x + 3a°x* — ax’ + a. 


at — 2a'xe + 8a°x? — Qax* + «rt | a’ —ax + x*—the root. 
a’ . 
tee 80s? 
—2a?r + a’x” 
Qa°x? — Qax*® + a* 
Qa’? — Qax® + a* 


Qa? — ax — the first divisor. 





Qa? — 2ax + x? = second divisor. 





For verification, multiply this root by itself thus: 
a—ar +2 
a—axr +27 
ati—a’x +an 
—aae +a’x? — ax’ 
cae ag? —az® + 2% 
at — Qa*x + 3a°x? — 2ax® + x*, which is the orig- 





inal polynomial. 

Ex. 2. Required the square root of a’ + 2ab + 2ac + b+ 2be + c’. 

Ex. 3. Required the square root of 102" — 10x? — 122° + 5a? + 
92° — 2 +-1. 32?@— Dery HX —/ 

Ex. 4. Required the square root of 42* ss. Sax? + 4a°x? + 168% 24 
16ab’x + 166%. 
: Ans. 22° + 2ax+ a0, 

Ex. 5. Extract the square root of a ba ae oh oF oe 
15a°b* — Gab? +. 6°. 


Ex. 6. Extract the square root of a A aa hy A220 


(271.) To extract the cube root of a polynomial. 
We already know that the cube of a+0 is a’ + 3a’) + 3ab?+ 6°. 


= 
bes 


e 


_. EVOLUTION OF POLYNOMIALS. 225 


If, then, the cube were given, and we were required to find its 
root, it might be done by the following method: _ 

When the cube is arranged according to the powers of one let- 
ter, we at once know from the first term a’, that @ must be one 
term of the root. If, then, we subtract its cube from the proposed 
polynomial, we obtain the remainder 3a°) + 3ad?+ 3°, which must 
furnish the second term of the root. : 

Now this remainder may be put under the form © 


(3a? + 3ab + 6°) x b; 


whence it appears that we shall find the second term of the root, 
if we divide the remainder by 3a’ + 3ab + 0°. But as this second 
term is supposed to be unknown, the divisor cannot be completed. 
Nevertheless, we know the first term 3a’, that is, thrice the square 
of the first term already found, and by means of this we can find 
the other part 0, and then complete the divisor before we perform 
the division. For this purpose, we must add to 3a’ thrice the 
product of the two terms, or 3ab, and the square of the second 


term of the root, or *. Hence we derive the following 


RULE FOR EXTRACTING THE CUBE ROOT OF A POLY- 
NOMIAL. 


(272.) Having arranged the polynomial according to the powers of 
some one letter, take the cube root of the first term, and subtract the cube 
from the given quantity. 

Divide the first term ofthe remainder by three times the square of 
the root already found, the quotient will be the second term of the root. 

Complete the divisor by adding to it three times the product of the 
two terms of the root, and the square of the second term. | 

Multiply the divisor thus increased by the last term of the root, and 
subtract the product from the last remainder. Proceed in the same 
manner to find the additional terms of the root. 

Ex. 1. Extract the cube root of a’?-++ 12a°+ 48a-+ 64. 

a +- 1207+ 48a + 64 | @+ 4,— the root. 
° a” * 
12a? + 48a + 64 | 3a° + 12a + 16 = the divisor. 
12a°+ 48a + 64 | 


Having found the first term of the root a, and subtracted its 


Eg 


‘ ‘ . pert. [oe a 
ee 


226 EVOLUTION OF POLYNOMIALS. 


cube, we divide the first term of the remainder, 12a’, by three 
times the square of a, that is, 3a’, and we obtain 4 for the second 
term of the root. We then complete the divisor by adding to it 
three times the product of the two terms of the root, which is 
12a, together with the square of the last term, 4, which is 16. 
AMultiplpie, then, the complete divisor by 4, ata sulbgehecinie the 
product from the last remainder, nothing is left. Hence the re- 
quired cube root is a+ 4. sh 
This result may be easily verified by multi plica tigate 
Ex. 2. Extract the cube root of a® — 6a> + 15at — 20a* + 15a" 
—6a+1. 
a® — 6a°2b 15a* — 20a° + 15a? — 6a 4°21 | a?’ — 2a+ 1 = the root 
a 
— 6a® +..15a* — 200° | 3at — 6a° + 4a? = the first divisor. 
— 6a°+ 12a*— 8a’ | 
3a* — 12a°+ 15a? — 6a+ 1) 3a*—12a° + 150’?— 6a+1 
3a* — 12a°+ 15a°—6a+1]} =second divisor. 


5 


We may dispense with forming the complete divisor according 
to the rule, if, each time that we find a new term of the root, we 
raise the entire root to the third power, and subtract the cube 
from the given polynomial. A 


Ex. 8. Required the cube root of x° + 6x° — 402° + 96x — 64. Eis NS 


Ex. 4. Required the ¢ube root of 182 Pay 36x” a War +8 + 322° 
+ a - 6a". | 
Ex. 5. Required the cube root of b& — 5B. Mi 35° — : ay 36, 


(273.) To extract any root of a polynomial. 

We already know that the nth power of a+ 0d is a* + na"—'b + 
other terms. ‘The first term of the root is, therefore, the nth root 
of the first term of the polynomial. Also, the second term of the 
root may be found by dividing the second term of the polynomial 
by na"—'; that is, the first term of the root raised to the next in- 
ferior power, and multiplied by the index of the given powers 
Hence we deduce the following 


245-7) 


org ~~ 


a 


~ 


EVOLUTION OF POLYNOMIALS. 227 


RULE FOR EXTRACTING ANY ROOT OF A POLY. 
NOMIAL. 


Having arranged the terms according to the powers of one of the let- 
ters, take the nth root of the first term for the first term of the required 

root. 
Subtract its power from the given quantity, and divide the first term 
of the remainder, by n times the (n— 1) power of this root ; the quo- 
tient will be the second term of the root. 

Subtract the nth power of the terms already found, from the given 
quantity, and using the same divisor, proceed in like manner to find 
the remaining terms of the root. 

Ex. 1. Required the fourth root of 16a*— 96a°xr + 216a’x? — 
216ax* + 812%. 


16a‘ — 96a°ax + 216a°x? — 216ax* + 812% | 2a — 3x2 = the root. 
16a* 

— 96a°x | 32a* — the divisor. 
16a* — 96a*x + 216a°x” — 216ax* + 812%. 


Here we take the fourth root of 16a*, which is 2a, for the first 
term of the required root; subtract its fourth power, and bring 
down the first term of the remainder — 96a°x. For a divisor, we 
raise the first term of the root to the third power, and multiply it 
by 4, making 32a°. Dividing, we obtain — 32 for the second term 
of the root. The quantity 24a — 3x being raised to the fourth 
power, is found to be equal to the proposed polynomial. 


Ex. 2. Required the fifth root of 32x° — 80x* + 802* — 40x? + 
102 — 1. 
Ans. 22 — 1. 
Ex. 3. Required the fourth root of 812° — 2162’ 4. 336x° — 56x* 


— 22427 + 64a + 16. 
Ans. 3x? — 2x — 2. 


(274.) When the index of the root to be extracted is a multiple 
of two or more numbers, we may obtain the root required by the 
successive extraction of simpler roots, Art. 159. 

For example, we may obtain the fourth root by extracting the 
square root ¢wice successively ; for the square root of a* is a’, and 
the square root of a’ is a. 

The eighth root may be obtained by extracting the square root 


$228 EVOLUTION OF POLYNOMIALS. 


three times successively ; ; for the square root of a® is a‘, that of a* 
is a’, and that of a’ is a. 

ta the same manner, the sixteenth root may be pened by ex- 
tracting the square root four times successively, and so on. 

The sixth root may be found by extracting the square root, and 
afterward the cube root; for the square root of a’ is a’, and the 
eube root of a isa. We may also take, first, the cube root, which 
gives a’, and afterward the square root, which gives a, as before. 
It is, however, best to extract the roots of the lowest degree first, 
because the operation is less laborious. 

In general, che mnth root of a number is equal to the nth root of the 
mth root of this number. ‘That is, 


mn nm grr 
a=y/ Va. 


For raising each member of this expression to the mth power, 
we have 


“la — W/ as 
Ex. 1. Find the fourth root of at — 4°} + 6a°b* — 4ab* + 0 
Ex. 2. Find the sixth root of a®°-+ 6a°d se 15a*b? + 4's + 15a7b* 
+ 6ab° + 6° 
Ex. 3. Find the eighth root of 256z* + o2dary 4 1792a°y? + 
1792a°y? + 1120a'y* + 4482°y? + 112a°y° + 16ay' + 7’. Ley 


EXTRACTION OF THE SQUARE ROOT OF A QUANTITY OF THE FORM 

at Jb. 

(275.) Binomials of this class require particular attention, be- 
cause they frequently occur in the solution of equations of the 
fourth degree, suchas are treated of in Art. 184. Thus, the equa- 
tion 

a* = 62° — 4, 
gives us D sey 10s 

Hence, in order to find the value of x, we must extract the 
square root of the binomial 3+ /5. _ 

In order to show that the square root of such an expression 
may sometimes be extracted, take the binomial 


2+ V3, 


and find its square. 


EVOLUTION OF POLYNOMIALS. 920% 


(24 J Mex oho 87+ 4 8. 


Therefore, the square root of 7+4/3 is 2+ V3. 

The square root of an expression of the form a+ V 6 may, 
therefore, sometimes be extracted, and it is required to determine 
a general method for this purpose whenever it is practicable. 


PROPOSITION I. 


The square root of a whole number cannot consist of two parts one 
of which is rational and the other irrational. 


For, if possible, let /a—x2+V y, where x denotes a rational 
quantity, and / y denotes a surd. 

By squaring both sides, we obtaina =a? + 22 /y+y; whence, 
by transposition and division, 


2 
a—nx—y 


skefaf=s On 


The second member of the equation contains only rational quan- 
tities, while /y was supposed to be zrrational ; that is, we find an 
irrational quantity equal to a rational one, which is absurd. 
Hence the square root of a whole number cannot be partly ra- 
tional and partly irrational ; in other words, the sum or difference 
of two surds cannot be equal to a whole number. 


PROPOSITION IL. 
In every equation of the form 
rt / y = (-f> J b, 


the rational parts on the opposite sides are equal to each other, and 
also the irrational parts. 


For if x is not equal to a, let it be equal to a+ z. 
Then ~ Gihebvysedk vb; 
or eA Bab 2 A yg 


that is, V6 is partly rational and partly irrational, which by the 
last Proposition is impossible. Therefore, r=a, and, consequent- 


ly, i J b:z 


= ye = 
ita 
~~ 
v's 
s 


 » 2380 EVOLUTION OF POLYNOMIALS. 


PROPOSITION III. 


If Vat Vb is equal tov + Vy, 


then will Va—vV/b be equal to x— Vy. 


For, by involution, a+ /b=2°?+ 2v7JVy+y. 
But by the last Proposition, a=2+y, 
and Jb = w/y. 


a— Jb=x2*— 24 JS y+y. 
Therefore, by evolution, /a— /b=r— Vy. 


Subtracting, we obtain, 


(277.) To find an expression for the square-root of at vb. 
Assume 


Ja+ vb =ptq (1), 


where p and g may be both radicals, or one rational and the other 
a radical, but p’ and q’ are necessarily rational. 
Then, by the last Proposition, — 


Va —Vb=p—q (2). 
Multiplying these equations together, we obtain 


Ja — b= p? —q’ (3), a rational quantity. 


Hence we see that, in order that\/a+,/0 may be expressed 
by the sum of two radicals, or one rational term and the other a 
radical, the expression a” — b must be a perfect square. 


Let, then, a? — 6 be a perfect square, and put /a®—b=c; 
equation (3) will thus become 
p— ore. 
Squaring equations (1) and (2), we obtain 
Pre + epg=at V6, 
is wi bles! dk Sree 
Adding these two equations, we obtain 


p + g° =. 
But we have already 
p —_— g iG. 
Hence Qn® =a +e, 


29° =a —c. 


ba 
< | 
a 
i 

. 





EVOLUTION OF POLYNOMIALS. Rad, 2 
) 2, 
ate 
And, therefore, p=t 
: a—c 
a <c ne 


Therefore, 


Jat Jb, opt+gq=Ht (E+ l=) 


Va— Jb, orp—qe+ (\/*#-V"=). 


(278.) Hence, to extract the square root of a binomial of the 
form a+ Vb, we have the following 





RULE. 


From the square of the rational part (a°), subtract the square of the 
arrational part (b); take the square root of the remainder, and calling 
that root c, the required root will be 


ate a—c 
\/ 2 +y/ a is 


Ex. 1. Required the square root of 4+2/3. 
Here a= 4, and /b)= 2/3; therefore, a —b=c? = 16 — 12 
=4; orc=2. Hence, by the above formula, the required root 





will be 


‘A + 2 fon 
AE [St av3t1 


Verification. 
The square of /3+1is 34+2/3+ 1—4-+4 2/3. 
Ex. 2. Required the square root of 11+ 6/2. 
Here a=11, and /b=6V/2; therefore, b=36 x2=72; and 
a’—b=49=c’. Hence c=7, and we find the square root of 
11+6/2is V9+ V2, or 3+ V2. 
Ex. 3. Required the square root of 11+ 2/30. 

; Ans. /64+ V5. 
Ex. 4. Required the square root of 2+ /3. 

Ans. /2+ v4. 


932 EVOLUTION: OF POLYNOMIALS. 


(279.) This method is applicable even when the binomial con- 
tains imaginary quantities, as in the following examples. 

x..). Required the square root of 144 / —3. 

Here a=1, and /b=4/ —3; hence 6 = —48, and a? —b= 
49; therefore,c—'7. The required square root is /44+ / —3 
=2+V7V —3. 3 

Ex. 6. Required the square root of —}+3/ —3. 

Ans. 4-year 

Ex. 7. Required the square root of 2./ — 1. 

Here we put a=o; hence c=—2, and the required root is 


1 dasha 
which may be easily verified. 


Ex. 8. Required the square root of 6+ 2/5 and 6 —2¥V5. 
» Ex. 9. Required the value of the expression 


\/443V— 20-44) 4—3 V0. 


Ans. 6. 


SECTION XVIII. 


INFINITE SERIES. 


(280.) AN infinite series is an infinite number of terms connected 
together by the signs + or —, and usually following:'some certam 
law. 


As t+4 
or 1} 

In the first of these examples, the terms are the reciprocals of 
the odd numbers, 1, 3, 5, 7, &c.; and in the other, they are the 
reciprocals of the even numbers, 2, 4, 6, 8, &c., with signs alter- 
nately + and —. 

Infinite series may arise from the common operations of divis- 
ion, the extraction of roots, and other processes of calculation, 
as will be seen hereafter. 

Al converging series is one in which the terms continually de- 
crease, as in the examples just given. 

Al diverging series is one in which the terms continually in- 
crease ; as, | 


129.4 48+ 16— $24) Se. 


/ln ascending series is one in which the powers of the unknown 
quantity continually increase ; as, 


1+ ax + bx? + ca*® + dx*+ ex’ +, &e. 


Al descending series is one in which the powers of the unknown 
quantity continually decrease ; as, 


1+ ax! +. bx? + ca? + da-* + e#-° +, &e 


234 INFINITE SERIES. 


PROBLEM I. 


(281.) Any series being given, to find its several orders of differ- 
ences. 


RULE. 


1. Take the first term from the second, the second from the third, the 
third from the fourth, &c. ; and the remainders will form a new series, 
called the FIRST ORDER OF DIFFERENCES. 

2. Take the first term of this last series from the second, the second 
from the third, the third from the fourth, &c. ; and the remainders will 
form a third series, called the SECOND ORDER OF DIFFERENCES. 

3. Proceed in like manner for the third, fourth, fifth, &c., orders of 
differences ; and so on till they terminate, or are carried as far as may 
be thought necessary. 


Ex. 1. Required the several orders of differences of the series 


of squares, P49 16%" 25° BG 749. Tee. 
BY De wtd he wt Ladd Le first differences. 
A hs Ee Pea second differences. 
hs treed th Peg third differences. 
Ex. 2. Required the several orders of differences of the series 
of cubes, 1 78) 27 ©6444 12 2 1bp cee: 
Tin kOoo. Syt wT 897 first differences. 
12; 11610 24th 30 second differences. 
6:8 507 a 6 third differences. 


0 -0 fourth differences. 


Ex. 3. Required the several orders of differences of the series 
of fourth powers, 


1, 16, 81, 256, 625, 1296, &e. 


Ex. 4. Required the several orders of differences of the series 
of fifth powers, 


1, 32, 243, 1024, 3125, 7776, 16807, &c. 


Ex. 5. Required the several orders of differences of the bean 


of numbers, 


1, 3, 6, 10, 15, 21, &e. 


Y Ove" | 


"INFINITE SERIES. 


to 
oo 
or 


PROBLEM II. 


(282.) To find the nth term of the series 
Oy le Gy dy &,' 8cC., 


when the differences of any order become at last equal to each 
other. 

Take the proposed series, and subtract each term from the next 
succeeding one; we shall obtain the following series, which is 


called the first order of differences : 
b—a, c—b, d—c, e—d, f —e, &e. 


Again, subtracting each term of this series from the next suc- 
ceeding term, we find for the second order of differences, 


c—2b+a, d—2c+b, e—2d+c, f—2%e+d, &e. 


Subtracting, again, each term of the preceding series from its 
next succeeding term, we get the third order of differences, 


d— 3c + 3b—a, e— 3d+ 3c —b, f— 3e+ 3d—c, Ke. 
Subtracting again, we find for the fourth order of differences, 
e—4d+ 6¢—4b+a, f—4e+ 6d—4c+), &e. 
And for the fifth order of differences, 
f —5e+ 10d — 10c + 5b —a, &c. 


Let D’, D’, D’’, D, &c., represent the first terms of the sev- 
eral orders of differences. 


Then, 
D’ =b—a; whence 6=a+ D’ 
DY’ =c—2b+a; whence c=a+2D’+ D” 


D’’ =d— 3c +3b—a; whence d=a+3D’+3D”+ D’”’ 
D* =e—4d+6c—4b+a; whence e=a+4D’+6D’+4D’’"+D*, 
&c., &e. 


The coefficients of the value of c, the ¢hird term of the proposed 
series, are 1, 2, 1, which are the coefficients of the second power 
of a binomial ; the coefficients of the value of d, the fourth term, 
are 1, 3, 3, 1, which are the coefficients of the third power of a 
binomial; and so on. Hence we infer that the coefficients of the 


236 INFINITE SERIES. 





ath term of the series, are the coefiicients of the (2 — 1) power 
of a binomial. Therefore, the mth term of the series will be 


(n—1),,, (n—1)(n—2),,, , (n—1) (n—2) (n--3)__, 
ae Dh eee ae: ee 
&e. . 
Ex. 1. Required the twelfth term of the series, 
Oy Comer ao0, oO, eos 


The first order of differencesis 4 6-8 10, &e. 
The second order of differencesis 2 2 2 &e. | 
The third order of differences is Ugh 


Therefore, D4, D’=2, and D”=0. Also, a=2, and n=12. 
nm — 1) (2 — 2)D” 
2 


! 
| 
| 
| 


=2+11D’+ 55D” = 





Then a + (2 —1)D’+ ( 

= 2+ 44+ 110 = 156 — the twelfth term. 
Ex. 2. Required the twentieth term of the series, 

Tj ty 005 LO, One Le aoc ee 

Here, recy i De?) bBo and n=20 
Therefore, the 20th term=1+19D’+ 171D” — 1+ 88+ 171 
= 210, Ans. 
Ex. 3. Required the thirteenth term of the series, | 


1, 2, 3, 6, 12, 24, 48, &c. rete 
Ans. —, fribow vevied 


Ex. 4. Required the fifteenth term of the series, 


IY 4, (9571859 255°36,""tcc. 
Ans. 225. 
Ex. 5. Required the twentieth term of the series, 


1, 8, 27, 64, 125, &e. . 000 ~< 


PROBLEM III. 
(283.) To find the sum of n terms of the series, 

a, b, c, d, e, &e. 
Assume the series, 


0, a,a+b,a+b+c, atb-+c+d, &e. 


INFINITE SERIES. 237 


Subtracting each term from the next succeeding, we obtain 
a, b, c, d, e, &c., 


which is the series whose sum we propose to find. Hence the 
sum of 2 terms of the proposed series, is the (x + 1)th term of 
the assumed series; and the mth order of differences in one series 
is the (x + 1)th order in the other series. If, therefore, in the 
formula of the preceding Problem, we substitute 
0 for a, ; 
n+1 for n, 
a for D’, 
19 et Cd Dp 
ite OCC, 
we shall have 
n(n—1),,, n(nm—1)(n—2),,, , n(n—1) (n—2) (n—3)_,,, 
ge Praag 23.4 q 
+, &e., 
which is the sum of 7 terms of the proposed series. 
Ex. 1. Required the sum of 2 terms of the series 
12, 3440 5. 6, Sc. 
Here | eal, 1) D0: 
n(n — 1)D’ 7 —§2 = m+n n+1 
Therefore, na + Boe =n2-+ a = — ay re pe 
the sum of 2 terms, the same as found in Art. 239, 
Ex. 2. Required the sum of 7 terms of the series 
1?, 2, 3%, 4°, 5%, &e. . 
Here Gz 1, 1) 3.3} Dito: 
Therefore the general formula reduces to 


meu Me 1) | 2n(m — 1) (n — 2) 


_ 208+ nin 


6 


__ m(n+1) (2n+4 1) 
aA 6 


“fe oe ei 


pp F 





» the sum required. 


938 INFINITE SERIES. 


Ex. 3. Required the sum of m terms of the series 
1°, 28, waa) D', 5, CC 


Here & alps wD i ye REA, 
Ans. A vw 
Ex. 4. Required the sum of » terms of the series ss 
153,205 10, 15; kee: 
ne n(n + 1) (n+ 2) 


2.3 
Ex. 5. Required the sum of 7 terms of the series 
1, Ayal, 205 30, dec. 
a(n + 1) (n+ 2) (n+ 8) 
Ans, ———_—____—, 
jal 2.3.4 
PROBLEM IV. ¢ 

(254.) Any series of equidistant terms, a, b, c, d, e, &c., being given, 
to find any intermediate term by interpolation. 

This is essentially the same as Problem II. For convenience, 
let us put z to represent the distance of the required term from 
the first term of the series a, in’which case 2 = 2 — 1, and we 
shall have / 

D’a(e2—1)  D’’a(2 — 1) (x — 2) 


z—-a+De+ 5 23 +, Sew 
Ex. 1. Given the square root of 160, equal to 12.64911; 
ws +S epi Hers 12-72 7923 
as oe 1640" 12.80625, 


to find the square root of 161. 


Here the first differences are +.07881, +.07833. 

And the second difference is —.00048. 

The interval between the given numbers is 2; the distance of 
the required term from the first term-is 1; and, since this is al- 
ways to be reckoned in parts of the equal intervals of the given 
numbers, we have « = . | 

Also, D’= + .07881, D” = — .00048. 

Hence z=a+1D’—1D’, 

= 12.64911 + .03941 + .00006, 
— 12.68858, which is the square root of 161. 


INFINITE SERIES. 239 


Ex. 2. Given the cube root of 60, equal to 3.91487 ; 


6 66 62, és 3.95789 ; 
«“ “ 64,  “ 4.00000; 
mie “ “ 66, 4.04124, 
to find the cube root of 61. 
; Ans. 3.93650. 
Ex. 3. Given the fourth root of 625, equal to 5.000000 ; 
« di 628, “ 5.005988; 
«“ «“ 631, “ 5.011956; 
«“ « 634, “ 5.017903, 


to find the fourth root of 627 


Here e=2. Therefore, z=a+2D’—1iD”. 
Ans. 5.003994. 


Ex. 4. Given the square root of 70, equal to 8.36660 ; 


« Cd ATA nil 8.60233 : 
« « Hse ese 18. 89776. 
“ « 82, * 9.05539, 


to find the square root of 71. 
Ans. 8.42615. 


(285.) Fractions expanded into infinite series. 
When the dividend is not divisible by the divisor, the quotient 
may be expressed bya fraction. Thus, if it is required to divide 


1 by 1—a, we obtain the fraction We may, however, 


yee 
proceed with the division according to the usual method, thus: 





1 1—a 
— a 1l+a+a-+a+a‘*-+, &c., = the quotient. 
Bays : 
a—a@ 
ae A 
ai — a 
a 
a> — a 
“pA tat 
1 ; : 
Hence gays 1+a+a@+a+at+a’4, &c., to infinity. 


24.0 INFINITE SERIES. 


Suppose a = 4, we shall then have 


1 1 


ee ; =2, which will be equal to the series 
1—a 1 — Dy 


1+3+h+4+75 +, &. 
Suppose a = 1, we shall then have 
1 1 











= 3, which will be equal to the series 


1+it}+a tart, &e. 





Ex. 2. Resolve into an infinite series. s 


1+ a 
Ans. 1— a+ a°—a’+ a'—a’+, &e. 


a 


Suppose a = 1, we shall then have 


Fy 


= 2, which will be equal to the series 
aa WORE Som or iat =o +, ee 


: Cie i , : 
Ex. 3. Resolve the fraction —-— into an infinite series. 





a+b 
c = OL nae 
Ans. see rea sb aang 
a ae ate axe Ae, 9 \ 
Ex. 4. Resolve Fly into an infinite,series. -j “gs 474s. wae Ne 


ih ies } ; , + 

Ex. 5. Resolve nee into an inate series.  /-+ 4x4 Xx 2p 2794-4 ae 
We may proceed in the same manner when there are more — 4 } 
| aa 


, than two ‘terms in the divisor. 


1 ; . Sale’ : 
Ex. 6. Resolve Tog rg into an infinite series 


—a-+a 


Ans. 14+a—a’— a‘ + a°+, &e. 
Ex. 7. Resolve ———, into an infinite series. 
(a sae way 
(286.) Infinite series obtained if extracting the square root. 


In Art. 267, /a-+ 65 has been expanded into an infinite series 
by the Binomial Theorem. It was also remarked that the same 


a in 


INFINITE SERIES. 241 


result might have been obtained by extracting the square root 
according to the usual rule. The opération will be as follows : 


b? b° 











1. 6 5b* 
a+b} a? tare ait aT = a T98ai +? &c.,= square root of a+ b. 
@ | 
‘GeO ae. 
b Qa% + —1, first divisor... 
Qa2 
Bale iat 
b+ 7 
b 1) ap b? 
—— | 2a? + -1— —3, second divisor. 
4a ay az S8az ' 
b? ? bt 


Mi4a Sa" | 64a" 
b° ae pee ie b? 


Sa" Shae jaa 2g Rr: oes {qi third divisor. 
lia bt b° 5° 
Sa | 16a 64a * 25608 


5O* b° b° 
~bha® * 64a' 2560" 

This result is the same as that obtained in Art. 267. 

Ex. 2. Extract the square root of 1 + 2. 
Rehr Lae) ORG 
Ans. 1+ 5—% 16 19g + &¢ 
Ex. 3. Extract the square root of@’+0. 2+; = 
Ex. 4, Extract the square root of at— 6. a’~S2— 5 oe 


METHOD OF UNKNOWN COEFFICIENTS. 


(287.) The method of unknown coefficients is a method for the 
development of algebraic expressions, by assuming a series with 
unknown coefficients, and afterward finding the value of these 
coefficients. This method is founded on the following 


THEOREM. 
If an equation of the form 
A+ Ba + Ca2’?+ Da? 4+, &c.) = A’+ Bir + C’x’? + D’2’ +, &e., 


must be verified by any value given to x, the terms involving the same 
powers in the two members are respectively equal. 


TT. - 


942 INFINITE SERIES. 


For, since this equation must be verified for every value of a, it 
must be verified when 2 =0. But, upon this supposition, 


1 pe 
Suppressing these two equal terms, we have 
Ba + Ca*+ Da? +, &c., = Bia + Oa? + D’a? +, &e. 
Dividing every term by x, we have 
B+ Cr+ Dz’+, &c., = B’+ C’a+ D’a’+, &c. 


Since this equation must be verified for every value of a, it 
must be verified when z= 0. But, upon this supposition, 


B= b’ 
In the same manner, we can prove that 


(omrred Baa 
D =2 D3 &e. 


(288.) In order to give some idea of this method, we propose 





Tas into a series arranged accord- 
ing to the powers of a. It is plain that this development is pos- 
sible, for we may divide the numerator by the denominator, as 
explained in Art. 285. 


Let us, then, assume 


to develop the expression 


1— zx pus 2 54 4 
Tig ee Oe + Dz* + Ex‘ +, &c., 
A, B, C, D, being coefficients independent of x, but dependent 
on the known terms of the fraction. 

In order to obtain the values of these coefficients, let us multi- 
ply both members of the above equation by 1 a xv, and we shall 
have 


i—vx=A+(A+B)r+(B+4 C)a’?+ (C+ D)a’?+(D + E)at +, &e. 


But, according to the preceding Theorem, the terms involving 
the same powers of # in the two members of the equation must 
be equal to each other. 





INFINITE SERIES. 943 


Therefore, A= (1, 
A+B=—1; hence B = —2. 
B+C= 0; + C—+2. 
C+D= 0; “* D=—2. 
D+E—...0;..%, B= +2, 
&c., &e. 


Substituting these values of the coefficients in the assumed 
series, we obtain 


1—~2z 
1+2 


(289.) The method thus exemplified is expressed in the fol- 
lowing 


= 1— 22+ 22° — 2x’ 4+ 2a* —, Ke. 


RULE. 


Assume a series with unknown coefficients for that which is required 
to be found ; then, having multiplied it by the denominator of the given 
fraction, or raised it to its proper power, find the value of each of 
these coefficients by equating the corresponding terms of the two expres- 


sions, or putting such of them as have no corresponding terms, equal 
to zero. 


. 1] . . e e 
Ex. 2. Expand the fraction Ton a oe an infinite series. 
1 « 
2. ES A Ee 2 3 4 
Assume {20s =A++Bze+ Ca’?+ Da’?+ Ea'+, &e. 


Multiplying by 1— 2x + 2’, we have 
1=A-+(B— 2A)za + (C — 2B + A)a? + (D — 20 + B)a’ + (E— 


2D + C)a*+, &e. 
Hence we must have 
A=1 


Ge SOB Anideis G) - FC OReee Oo 
D—2C+B=—0 ‘-D=2C—B=4, 
&c., &e. 


Therefore, 7 


a diary eran a 
Jpg = 1+ et 32° + 4a* + Batt, &e. 


2 hh INFINITE SERIES. 


1+ 2x 
1—2— 
Ans. 1+ 3x + 4a? + 723 + 1iz* + 182° + 292°+, &c., 


where the coefficient of each term is equal to the sum of the 

coefficients of the two preceding terms. 
1—z 

Ex. 4. Expand cay gg 

Ans. 1+ 2+ 5a’ + 132° 4+ 41a*+ 1212°+, &e. 

1 + x 

1— 3x 

Ans. 1+ 52+ 152? + 4527+ 1352*+, &e. 


Ex. 6. Expand /1— 2 into an infinite series. 


ane x x 3a" 3.52% 3.5.7 xv 
8 TDi Bab 9.4 6.8 0) O46 eee 


(290.) The method of unknown coefficients requires that we 
should know beforehand the form of the development, with re- 
spect to the powers of x. Generally, we suppose the develop- 
ment to proceed according to the ascending powers of x, com- 
mencing with 2°; but sometimes this form is znapplicable, in which 
case the result of the operation is sure to indicate it. 

Let it be required, for example, to develop the expression 


Ex. 3. Expand the fraction = into an infinite series. 


into an infinite series. 


into an infinite series. 





Ex. 5. Expand 


&e. 


Po into a series. 





Assume ae soy = A-+ Br + Ca’+ Da? +, &c. 


Clearing of fractions, we have 
; 1 = 3Az + (8B — A)z’ + (8C — B)a*+, &e. ; 
whence, according to Art. 287, we conclude pi 
Le 0, 
3A = 0, &c. 
Now the first equation, 1 = 0, is absurd, and shows that the 


assumed form is not applicable in the present case. But if we 


; 1 
put the fraction under the form = A 





ha and suppose that 


1 | 


: 2 3 
aX 3g =z At Bet Cz + Dz +, &c.), 


oa 


INFINITE SERIES. 245 


it will become, after the reductions are made, 
1= 3A + (3B — A)z + (38C — B)az’ + (83D — C)a’ +, &e., 
which gives the equations 


3A =1; whence A= 
3B —A=0; har as. 
30_—B—0; A. FE 
3D— C=0; ee D 
+ 


1 Voy (8 aa eR ap 
Therefore, 37 — a2? — (3 + : 4+ Q7 
aa} 


— = +5 af a+ +, &e, 


that is, the development contains a term affected with a negative 
exponent. 


We ought, then, to have assumed at the outset 


3p ~ AT + B+ Cx + Da + Ea’ +, &c. 

The particular series which is to be adopted in each case may 
be determined by putting 2 = 0, and observing the nature of the 
result. If, in this case, the expression to be developed becomes 
equal to a finite quantity, the first term of the series will not con- 
tain z. If the expression reduces to zero, the first term will con- 

, A 
tain x ; and if the expression reduces to the form 0” then the 
first term of the development must contain 2 with a negative ex- 
ponent. 


SECTION XIX. 


GENERAL THEORY OF EQUATIONS. 


(291.) Ir is proposed in this Section to exhibit the most im- 
portant propositions relating to the theory of equations, together 
with the Theorem of Sturm, by which we are enabled to deter- 
mine the number of real roots of an equation. | 

A function of a quantity is any expression involving that quan- 
tity. Thus, 

ax’? + 6 is a function of a. 
ay® + cy+d is a function of y. 
ax?—by? is a function of x and y. 

In a series of terms, two successive signs constitute a perma- 
mence when the signs are alike ; and a variation when they are 
unlike. Thus, in the polynomial 

a+b—c-+d, 
the signs of the first two terms constitute a permanence ; the 


signs of the second and third constitute a variation ; and those 
of the third and fourth also a variation. 


(292.) A cubic equation is one in which the highest power of 
_ the unknown quantity is of the third degree, as, for example, 
x’ — 627+ 8x — 15 —0. 
All equations of the third degree may be reduced to the form 
a+ az’?t+ ber+c=—0. 

A biquadratic equation is one in which the highest power of the 

unknown quantity is of the fourth degree ; as, for example, 
a*—6a* + Tar? + 52 —4=—0. 


GENERAL THEORY OF EQUATIONS. 24:7 


Every equation of the fourth degree may be reduced to the 
form 


aan +betertd=0. © | 
The general form of an equation of the fifth degree is 
xv’ + aut + bz? + cx? +dr+e=0; 
and the general form of an equation of the mth degree is 
ot. Ag*—' + Ba 784. Co®-Ft 1... +Tz+V=0 (m). 


This equation will be frequently referred to hereafter by the 
name of che general equation of the mth degree, or simply by the 
letter (m). 

It is obvious, that if we could solve this equation, we should 
have the solution of every equation which could be proposed. 
Unfortunately, no general solution has ever been discovered ; 
yet many important properties are known which enable us to 
solve any numerical equation which can ever occur. 


PROPOSITION I. 


(293.) If a ts a root of the general equation of the mth degree, the 
equation will be exactly divisible by x —a. 

For if a is one value of a, the equation must be verified when 
we substitute a in the place of 7. Hence we must have 


weep Ba" s' + Bas teh Catt sf Cfre +Ta+V=0 (1). 
Subtracting equation (1) from equation.(m), we obtain 
(2™— a”) + A(v”—'’—a”—*) 4 B(x” ?— a" —*) +... +. T(w— a) =0 (2). 


But, by Art. 76, each of the expressions (x” —a”), (x"—~'—a”™—"), 
&c., is divisible by «—a; and therefore equation (2) is also 
divisible by e—a. Now equation (m) is essentially the same as 
equation (2), for if we take the value of V, as found from equation 
(1), and substitute it for V in equation (m), it will give us equa- 
tion (2); therefore, equation (m) is divisible by z— a. 

Conversely, if equation (m) is divisible by x —a, then a isa 
root of the equation. 

It will be noticed that this property is but a generalization of 
what has been proved of equations of the second degree in Art. 
192. 


248 GENERAL THEORY OF EQUATIONS. 


Ex. 1. Prove that 1 is a root of the equation 
a* — 62+ lle —6=0. 
This equation is divisible by x — 1, and gives x’ — 5r+6=0. 
Ex. 2. Prove that 2 is a root of the equation 
x —xr—6=—0. 
This equation is divisible by « — 2, and gives 2° + 2a+3=0. 
Ex. 3. Prove that 2 is a root of the equation 
x~ 2) a? Tia? + 362 — 36 = 0. (4°74 Ve 
Ex. 4. Prove that 4 is a root of the equation 
a + 2? — 34x +56 =0. (x* + 3? 
Ex. 5. Prove that —1 isa root of the “aenkstin 
| at — 380° + 2102" + 536r + 289 —0./ 29-374" F47 
Ex. 6. Prove that —5 es a rodt of the equation — | 
x’ + 62* — 10x* — 112%? — 20%xr — 110 —0. 
Ex. 7. Prove that 3 is a root of the equation 


a’ + a — 140° — 140 + 492° + 492? — 362 — 36 —0. ~ 


new 
; = 


PROPOSITION IL. 


(294.) Every equation containing but one unknown quantity, has 
as many roots as there are units in the highest power of the unknown 
quantity, and no more. 

For, suppose a to be a root of the general equation of the mth 
degree. By the last Proposition, this equation is divisible by 
x —aj; and if we actually perform the division, the equation will 
be reduced to one of the next inferior degree, and may be repre 
sented by the form 


ob hg" A Be" 4. Pet W=0 


This equation must also have a root, which we will represent 
by b; and dividing by x—4, the equation will be reduced to one 
of the next inferior degree, and so on. 

We may continue this operation (m— 1) times, when we shall 








—_ 


GENERAL THEORY OF. EQUATIONS. 249 


arrive at a simple equation which has only one root. Hence, the 
proposed equation will have m roots, 


a, b, C, d, e ® @4@ @ GC; 


and its successive divisors, or the factors of which it is composed, 
will be ; 
x—a,x—b,x—c,x—d,.....2—l, 


being equal in number to the units contained in m, the index of 
the highest term of the equation. 

We have seen that when one root of an equation is known, the 
depressed equation containing the remaining roots is readily 
found by division; and if we can depress any equation to a qua- 
dratic, its roots can be determined by methods already explained. 


Ex. 1. One root of the equation 
xv + 32? — 16z+ 12—0 


Fal 


is 1. Find the remaining roots. < 
Ex. 2. Two roots of the equation 


x* — 25x? + 607 — 36 = 0 
are land 3. Find the remaining roots. 
Ex. 3. Two roots of the equation 
ax* — 122° + 482? — 682 + 15—0 
are 3and 5. Find the remaining roots, 
Ex. 4. Two roots of the equation 
4a* — 142° — 52? + 3lz+6=0 
are 2 and 3. Find the remaining roots. 
Ex. 5. Two roots of the equation 
x *— 62° + 24a — 16 —0 
are 2 and —2. Find the remaining roots. 


(295.) It should be observed that this Proposition only proves 
that an equation of the mth degree may be continually depressed 
by division, and finally exhausted after m operations. The divi- 
sors are not necessarily unequal. Any number, and indeed all of 
them, may be equal. When we say that an equation of the mth 
degree has m roots, we mean that the polynomial can be decom- 


G, 
250 GENERAL THEORY OF EQUATIONS. 


posed into m binomial factors, equal or unequal, each containing 
one root. Thus, the equation 


xv? — 62? 4+ 122 —8=0 
can be resolved into the factors 
(x — 2) (a — 2) (wv —2)=03; or (x — 2)7=0; 
whence it appears that the three roots of this equation are 
2, 2, 2. 


But, in general, the several roots of an equation differ from 
each other numerically. 
The equation 


2? —8 


has apparently but one root, viz., 2; but by the method of the 
preceding article we can discover two other roots. Dividing 
x —8 by x —2, we obtain 


x? + On +4—0. 
Solving this equation, we find 
| ct=—1lti J —3 
Thus, the three roots of the equation z* = 8 are 
235 — 1+ Ve 8s el] Vf 28. 


These last two values may be verified by multiplication as 
follows: | 








hal es yide yf Sa) aidece, Vise 
nl es AN AE) hy Yel ey ae) 
ea) eee 14+ V—3 
ee P/O ey a on ae 
—2—2/—3=thesquare. —2+2/—3 =the square. 
a es fs wet Lie, fH 3 
24+2/—3 2—2/ —3 
my ra es ay ME We PRLS ry 


8 — the cube. 8 = the cube. 


GENERAL THEORY OF EQUATIONS. 251 


If the last term of an equation vanishes, as in the example 
xt + 2a* + 3x’ + 62 = 0, 


the equation is divisible by x —0; and consequently 0 is one of 
its roots. 
If the last two terms vanish, then two of its roots are equal to 0. 


PROPOSITION III. 


Law of the coefficients of every equation. 


(296.) In order to discover the law of the coefficients, let us 
form the equation whose roots are 


GUS Cl OAL, cig q.; 


This equation will contain the factors (2 — a), (x — 6), (e—c), 
&c.; that is, we shall have 


(x— a) (x —b) (x—c) (a@—d)..... (x —7)'=0. 


If we perform the multiplication as in Article 259, we shall 
have 


a2” —a\a2"—'+ ab| x—? —abc|x"—*§ +.....—(abe.....1) =0. 


—b + ac —abd 
—cC +ad —acd 
—d + be — bcd 
&e + bd &c. 
+ cd 
&c. 


Hence we perceive, 


1. The coefficient of the second term of any equation is equal to the 
sum of all the roots with their signs changed. 

2. The coefficient of the third term is equal to the sum of the prod- 
ucts of all the roots taken two and two. 

3. The coefficient of the fourth term is equal to the sum of the prod- 
ucts of all the roots taken three and three, with their signs changed. 

4. The last term is the product of all the roots with their signs 
changed. 

It will be perceived that these properties include those of qua- 
dratic equations mentioned on page 163. 

If the roots are all negative, the signs of all the terms of the 


252 GENERAL THEORY OF EQUATIONS. 


equation will be positive, because the factors of which the equa 
tion is composed are all positive. 

If the roots are all positive, the signs of the terms will be al- 
ternately + and —. 

Ex. 1. Form the equation whose roots are 1, 2, and 3. 

For this purpose, we must multiply together the factors # — 4, 
x — 2, x — 3, and we obtain 

xv’ — 62? + 1lz—6=0. 

This example conforms to the rules above given for the coeth- 
cients. Thus, the coefficient of the second term is equal to the 
sum of all the roots (1+ 2+ 3) with their signs changed. 


The coefficient of the third term is the sum of the products of 
the roots taken two and two; thus, 


1x24+1x3+4+2~x3. 


The last term is the product of all the roots (1 x 2 x 3) with 
their signs changed. 
Ex. 2. Form the equation whose roots are 2, 3, 5, and —6. 
Ans. x* — 4x? — 29a? + 1567 — 180 — 0. 


Show how these coefficients conform to the laws above given. 
Ex. 3. Form the equation whose roots are 1, 3,5, — 2, —4, —6. 
Ans. 2° + 32° — 412* — 872? + 4002? + 4442 — 720 —0. 


(297.) Every rational root of an equation is a divisor of the last 
term ; for, since this term is the product of all the roots, it must 
be divisible by each of them. If, then, we wish to find a root by 
trial, we know at once what numbers we must employ. 


For example, take the equation 
x—x—6=0. 


If this equation has a rational root, it must be a divisor of the 
last term 6; hence we must try the numbers 1, 2, 3, 6. 


If x=1,wehave 1—1—6=—6, 
Deas wes “4g He 8 —2—6—= 0, 
i) oma a dani 1s 27—3—6= 18, 
2a=—6,% “* 216—6—6= 204, 


Hence we see that 2 is one of the roots of the given equation, _ 
, 


GENERAL THEORY OF EQUATIONS. 253 


and by the method of Art. 294, we shall find the remaining roots 
to be 


—liv —2. 


PROPOSITION IV. 


(298.) No equation whose coefficients are all integers, and that of 
the highest power of the unknown quantity unity, can have a fractional 
root. 


For, take the equation 


v+Av’+Be+C=0, 


:' an fe-f 
and suppose, if possible, that the irreducible fraction Fe iS one 


value of x. If we substitute this value for z in the given equa- 
tion, we shall have 


a ax a 
prAgpts; +C=0. 
Multiplying each term by 2’, and transposing, we obtain 


aia (Aa? + Bad + Co”). 
Now by supposition, A, B, C,a and 4 are whole numbers. Hence 
the entire right-hand member of the equation is a whole number. 
But by hypothesis, is an irreducible fraction ; that is, a and J 
contain no common factor. Consequently, a® and d will contain 
3 


. ° a ° « ° e 
no common factor, that is, 7 is a fraction in its lowest terms. 


) wa 
Hence, if , Were a root of the proposed equation, we should have 


a fraction in its lowest terms equal to a whole number, which is 
absurd. 

The same mode of demonstration is applicable to the general 
equation of the mth degree. 


24 GENERAL THEORY OF EQUATIONS. 


PROPOSITION V. 


(299.) If the signs of the alternate terms in an equation be changed, 
the signs of all the roots will be changed. 

If we take the general equation of the mth degree, and change 
the signs of the alternate terms, we shall have — 


ao” — Ac*—14 Bo -*—_ Cn" 405... =0 (1); 
or, changing the sign of every term of the last equation, 
— 7" Ag”) Ba? + Ozr" fe .  Y ect 8 be A) 


Now, substituting + a@ for x in equation (m) will give the same 
result as substituting —a in equation (1), if m be an even number ; 
or, substituting —a in equation (2), if m be an odd number. If, 
then, a is a root of equation (m), —a will be a root of equation 
(1), and of course a root of equation (2), which is identical with it. 

Hence we see that the positive roots may be changed into 
negative roots, and the reverse, by simply changing the signs of 
the alternate terms; so that the finding the real roots of any 
equation is reduced to finding positive roots only. 

Ex. 1. The roots of the equation 

x* — 22°? — 5x+6=0 
are 1, 3, and —2. What are the roots of the equation 
x + 22° — 5a—6=01 af , Te 
Ex. 2. The roots of the equation 
a*— 677+ lla —6=0 


are 1, 2, and 3. What are the roots of the equation 


a + 62? + llz+6=—01 He a TT ee ee 


PROPOSITION VI. 


(300.) If an equation whose coefficients are all real, contains imagi- 
nary roots, the number of these roots must be even. 

If an equation whose coefficients are all real, has a root of the 
form | 


a+tb/—l1, 
then will a—bh/—1 


be also a root of the equation. 


GENERAL THEORY OF EQUATIONS. 255 


For, from Art. 296, the product of all the roots must be equal to 
areal quantity. Nowthe only factor which will render a+b/ —1 


real, is a— b./ — 1, or ac—beV —1. But the sum of all the 
roots must also be equal to a real quantity ; therefore the imagi- 


nary parts of the two roots must be equal, and have opposite 
signs. 


Ex. 1. Find the roots of the equation 
e— 2r+4—0. 
Ans. —2,14+/ —1. 
Ex. 2. Find the roots of the equation _ 
ee —a— Tr +.15—0. 
Ans. —3,2+/ —L 
Ex. 3. Find the roots of the equation 
52° + 22 —44—0. 
Ans. 2,—1ivJd —34. 


PROPOSITION VII. 


(301.) Every equation must have as many variations of sign as tt 
has positive roots ; and as many permanences of sign as there are neg- 
ative roots. 

To prove this Proposition, it is only necessary to show that the 
multiplication of an equation by a new factor, e—a, correspond- 
ing to a positive root, will introduce at least one variation ; and 
that the multiplication by a factor 2 + a@ will introduce at least 
one permanence. | 

For an example, take the equation 


a + 32? — 102 — 24=—0, 
in which the signs are + + ——, giving one variation. 
Multiply this equation by « — 2= 0, as follows: 
e+ 327— 107 — 24 
xz —2 
a + 382° — 102? — 24a 
— 2%? — 6a’? + 20x + 48 
att a —162°— 4c +48=0. 





 « 


256 = ss @ENERAL THEORY OF EQUATIONS. 


In this last product the signs are + + —— +4, giving two va- 
riations ; that is, the introduction of a positive root has intro- 
duced one new variation in the signs of the terms. — 

To generalize this reasoning, we perceive that the signs in the 
upper line of the partial products are the same as in the given 
equation ; but those in the Jower line are all contrary to those of 
the given equation, and advanced one term toward the right. 

Now, as long as each coefficient of the upper line is greater 
than the corresponding one in the lower, it will determine the 
sign of every term of the product. Hence, in this case, there will 
be in the product, with the exception of the last term, the same 
changes of sign as in the given equation. But the last term in- 
troduces a new variation, since its sign is contrary to that which 
immediately precedes it. 

When a term in the lower line is larger than the corresponding 
one in the upper line, and has the contrary sign, there is a change 
from a permanence to a variation; for the lower sign is always 
contrary to the preceding upper sign. Hence, whenever we:are 
obliged to descend from the upper to the lower line in order to 
determine the sign of the product, there is a variation which is 
not found in the proposed equation; and as all the remaining 
signs of the lower line are contrary to the preceding ones of the 
upper line, there must be the same changes of sign in this line as 
in the proposed equation. If we are obliged to reascend to the 
upper line, we may suppose the result to be either a variation of 
a permanence. But even if it were a permanence, since the last 
sign of the product is in the lower line, it is necessary to go once 
more from the upper line to the lower, than from the lower to 
the upper. Hence each factor, corresponding to a positive root, 
must introduce at least one new variation ; so that there must be 
as many variations as there are positive roots. 

In the same manner, we may prove that the multiplication by a 
factor x + a, corresponding to a negative root, must introduce at 
least one new permanence ; so that there must be as many perma- 
nences as there are negative roots. 


Ex. 1. The roots of the equation 
wv — 3x* — 5a? + 152? + 4a — 12 = 0 


are 1, 2,3, — 1, and —2. There are also three variations of sign, 


‘s 


GENERAL THEORY OF EQUATIONS. 257 


and two permanences, as there should be, according to the Prop- 
Osition. 

Ex. 2. The equation 

xt — 32° — 152? + 492 —12—0 

has four real roots. How many of these are negative 4 

Ex. 3. The equation 

x + 32° — 41a2* — 87x’ + 4002? + 4442 — 720 =0 

has six real roots. How many of these are positive 2 

If all the roots of an equation are real, the number of positive 
roots must be the same as the number of variations, and the num- 
ber of negative roots must be the same as the number of perma- 
nences. 


PROPOSITION VIII. 


(302.) Any equation may be transformed into another, whose roots 
shall be greater or less than those of the former by any given quantity. 

Let it be required to transform the general equation of the mth 
degree into another whose roots shall be greater by r than those 
of the given equation. 





Take y=u+r, orx=y—T, 
and substitute y— r for x in the proposed equation; we shall have 
—1)r| — 1)(m—2)r° 
y"—mr ial + mm 5} r yr mim see r oa, &c., — OQ 
Woe) pea! 2 
+A —(m— 1)Ar + fied) u ES 
: +B — (m—2)Br 
+C 


which equation evidently fulfils the required conditions, since 


is greater than a by r. 
If we take y=x2—r, or e=y-+7, we shall obtain in the same 


way an equation whose roots are Jess than those of the given 
equation by 7. 

Ex. 1, Find the equation whose roots are greater by 1 than 
those of the equation | 
xv’ + 327 — 4a +1—0. 

Ans. y°-— Ty +7=0 
R 


a” joie a) ad = is oe ee 


7 


| J 


* es. + GENERAL THEORY OF EQUATIONS.. 
« a a 


_ Ex. 2. Find the equation whose roots are less by 1 than those 
of the equation 


; x — 22°+ 32 —4=0. 
*:, Ans. y + y+ 2y— 2=0. 
* - Ex. 3. Find the equation whose roots are greater by 3 than 
those of the equation 


has a* + 92° + 122° — 14v — 0. 
“2 Ans. yt — 3y’— 15y? + 49y — 12 =0. 


Ex. 4. Find the equation whose roots are less by 2 than those 
of the equation ' 


Hat — 199° 4+- 32°+- 42 — 5 —0. 
Ans. 5y‘ + 28y° + 51y* + 32y —1=0. 


Ex. 5. Find the equation eres roots are less by 2 than those 
of the equation 


a + 22°—6a° — 102+ 8=0. 
Ans. y+ 10yt + 42y° + 86y? + My + 12=0. 


ee ky PROPOSITION IX. 


; aq 
(303.) Iny complete equation may be transformed. into another 
whose second term ts wanting. 
Since r in the preceding Proposition is indeterminate, we may 
put —mr-+ A equal to zero, which will cause the second term 


: A 
of the general development to disappear. Hence r= es and 


——— vy a = 
aaa 
Hence, to remove the second term of an equation, substitute for 
the unknown quantity a new unknown quantity, together with such a 
part of the coefficient of the second term, taken with a contrary sign, as 
BF denoted by the degree of the equation. 


oonBx. 1. Transform the equation Os 


. : Th 7 ae ee 
Y ta x c . t : a mm: . . 
FSS, —Ge'+82—2=0 eae 
~ > se ‘ey 
into another whose second term is wanting. ee 


yy Here we take a new unknown quantity, and annex to it a third 


$3 


? - ~ - ; % ' 


“? : 4 ) wis 





af Oe eee a 


GENERAL THEORY OF EQUATIONS. 959 


part of the coefficient of the second term of the equation with its 
sign changed ; that is, we putx=y+2. Making this substitu- 
tion, we obtain 
y — 4y—2=0, Ans. 
Ex. 2. Transform the equation 


zt — 162°— 62+ 15=—0 
into another whose second term is wanting. 
Here we put c=y+4. 
Ans. y* — 96y? —518y — 777 =0. 
Ex. 3. Transform the equation | 
x + 15x* + 120° — 20x” + l4a — 25—0 


into another whose second term is wanting. 
Ans. y° — 787? + 412y’ — T57y + 401 = 0. 


PROPOSITION X. 


(304.) If two numbers, when substituted for the unknown quantity 
in an equation, give results with contrary signs, there must be at least 
one root comprised between those numbers. 

Take, for example, the equation 


xv — 2x? + 32 —44—0. 


If we substitute 3 for x in this equation, we obtain — 26; and 
if we substitute 5 for x, we obtain +46. There must, therefore, 
be a real root between 3 and 5; for, when we suppose x = 3, 


x + 34 < 2x? + 44, 
But when "6 suppose 7 = 5, 

xv + 3x > 2x*?+ 44. 
Now both the quantities 

x +. 37 and 2a’+ 44 


increase while x increases. And since the first of these quanti- 
ties, which was originally less than the second, has become the 
greater, it must increase more rapidly than the second. There 
must, therefore, be a point at which the two magnitudes are 


260 GENERAL THEORY OF EQUATIONS. 


equal, and that value of x which renders these two magnitudes 
equal must be a root of the proposed equation. 

In the same manner, it may be proved that zf any quantity p, and 
every quantity greater than p, substituted in an equation, renders the 
result positive, then p is greater than the greatest root. 

Hence also, if the signs of the alternate terms are changed, and 
af q, and every quantity greater than q, renders the result a i then 
— q is less than the least root. 

If the two numbers, which give results with contrary signs, dif- 
fer from each other only by unity, it is plain that we have found 
the integral part of the root. 

Ex. 1. Find the integral part of one of the roots of the equation 


Qa*— 1lz’? + 82 — 16 —0. 


When a = 2, the equation reduces to — 12; and when z = 38, 
it reduces to + 71. Hence there must be a root between 2 and 
3; that is, 2 is the first figure of one of the roots. 

Ex. 2. Find the first figure of one of the roots of the equation 


x +o? +r — 100—0. 
Ans. 4. 


Ex. 3. Find the first figure of one of the roots of the equation 
ve — 4a? — 672 +80. 


w 


PROPOSITION XI. 


(305.) Of Derived Polynomials. 
If we substitute y + 7 for 2 in the general equation of the mth 
degree, we shall find the coefficients of r follow a remarkable law. | 

The equation, before it is developed, will be 


(ytr)”*+A(ytr)*' + Biytr)*?+..... + Tiytr)+V=0. 


If we actually involve the several terms (y + 7)”, (y + r)"—}, 
&c., as was done in Art. 302, we shall obtain certain terms inde- 
pendent of r, others which contain the first power of 7, others the 
second power of r, and so on; and the development will be of 
the following form : 

X 


X 
Pere} TC? ek S004 oat m 
Cp LBiie) 2.3.4 ee a 





GENERAL THEORY OF EQUATIONS. 261 


where the values of X, X,, X, &c., are 


eee + Ay"! + Bytae Cyr 8 ek fe aan» + Ty+V. 
X, =my”"~' + (m— 1)Ay”"~? + (m— 2)By”™—? + ..... + 28y + T. 
X,, =m(m — 1)y"—? + (m— 1) (m— 2)Ay”™—? +... 


Each of these polynomials may be derived from that tmmediately pre- - 
ceding it, by multiplying each term by the index of y in that term, and 
diminishing the index by unity. 

The expressions X, X_, &c., are called derived polynomials 
of X. X is called the first derived polynomial, X,, the second de- 
rived polynomial, X,, the third, and so on. 


Ex. 1. Find the equation whose roots are less by r than those 
of the equation 


2’ —5r2+6—0. 


Here we shall have 


) 


xX = y — 5y +6, 


X, caer 2y ee 5, 
>. sas 2, 
XK i — 0. 


But we have seen that when y +7 is substituted for zx, the 
equation reduces to the form 


ep gaara 2 mee oF &e. 





Substituting the values of X, x, X,, &¢e., above found, we ob- 
tain 


Ca pee gah 9 Fareh C2 orb ln hm 
which is the development of 
(y+r)?—5y +7) +6. 


Ex. 2. Find the equation whose roots are less by 7 than those 
of the equation 


xv’ — Tx’? + 8r—3=0. 


Here we shall have 


* 962 GENERAL THEORY OF EQUATIONS. 


X = y— Ty’'+8y—3, 
X =3y'—14y +8 


xX, > by ory 14, 
Se, a) Sg 
A 


and, substituting these values in the same formula as above, we 
obtain 


(y® — Ty? + 8y — 3) + (3y? — 14y 4+ 8)r+ : (6y — 14)r? + 5 67°, 
which is the development of | 
(y try —Ty +r? + 8y +7) —3. 
Ex. 3. Find the successive derived polynomials of the equation 
xt*— 82? + 14a? + 4¢—8—0. | 
Ex. 4. Find the successive derived polynomials of the equation 


x + 327 + 22° — 3x? — 22 —2—0. 


PROPOSITION XII. 


(306.) Of equal roots. 
We have seen, in Art. 295, that an equation may have two or 
more equal roots. Thus, the equation 


xv’ — 62°+ 122—8=0, 
or (e— 2)? =—0, 


has the three equal roots 2, 2,2. Such an equation and its first 
derived polynomial always contain a common divisor ; for the 
first derived polynomial of the above equation is 


32° — 122+ 12, 
or 3(a — 2)”, 


where it is evident that (e— 2)’ is a common divisor of both 
equations. And, generally, if the given equation and its first de- 
rived polynomial have a common divisor of the form (2 —a)”, the 
equation will have (m + 1) roots, each equal to a. 

‘To determine, therefore, whether an equation has equal roots, 
find the greatest common divisor between the equation and its first de- 


GENERAL THEORY OF EQUATIONS. 263 


ried polynomial. If there is no common divisor, the equation has no 
equal roots. If there is a common divisor, this must be a function of 
x. We then solve the equation formed by putting this equal to zero, 
and thus obtain all the equal roots. 


Ex. 1. Find the equal roots of the equation 

ax’? — 82? + 212 — 18 —0. 

The derived polynomial of this equation is 
3x? — 16x 4+ 21. 


The greatest common divisor between this and the given equa- 
tion is 
x — 3. 
Hence the equation has two roots, each equal to 3. 
Ex. 2. Find the equal roots of the equation 


x — 1327+ 552 — 75 — 0. 
Ex. 3. Find the equal roots of the equation 

ae — Tx? + 162 —12=0. 
Ex. 4. Find the equal roots of the equation 


x*— 6x? —8x—3=—0. 


PROPOSITION XIII. 


(307.) To find the number of real and imaginary roots of any equa- 
tion. 

In 1829, M. Sturm discovered a theorem which determines the 
precise number of real roots, and of course the number of imagi- 
nary ones; since the real and imaginary roots are together equal 
in number to the degree of the equation. We propose now to 
develop this theorem. 

Let X represent the first member of the general equation of the 
mth degree, which we suppose to have no equal roots, and let X, 
be its first derived polynomial, found by the method of Art. 305. 

Divide X by X, until the remainder is of a lower degree than 
the divisor, and call this remainder — X,,; that is, let X,, desig- 
nate the remainder witha contrary sign. Divide X, by X,, in the 
same manner, and so on, designating the successive remainders 


264 GENERAL THEORY OF EQUATIONS. 


with contrary signs by X,,,, X,,,, &c., until the division terminates 
by leaving a numerical remainder independent of x; which must 
always be the case, according to the preceding Proposition, since 
the equation having no equal roots, there can be no factor, which 
is a function of x, common to the equation and its first derived 
polynomial. Let this remainder, having its signs changed, be 
called X,,. 
The operation thus described will stand as follows: 





bet X, Xx, ey, “ | Xi 
x Pee! he “es | Q, Vi VQ 2, | Q, a GELS Qi 
+e X,Q,= Saat >a ) Mes AQ; = as ’ Aan — Nae = ae 


We thus obtain the series of quantities 


Rapky, Cie KX 


TR) “4 Lh) 


eee 


each of which is of a lower degree with respect to x than the 
preceding, and the last is altogether independent of z, that is, 
does not contain z. 

We now substitute for 2 in the above functions any two num- 
bers p and q, of which p is less thang. The substitution of p will 
give results either positive or negative. If we only take account 
of the signs, we shall obtain a certain number of variations anda 
certain number of permanences. 

The substitution of q for x will give a second series of signs, 
presenting a certain number of variations and permanences. The 
following, then, is 


THE THEOREM OF STURM. 


The difference between the number of variations of the first row of 
- signs and that of the second, is equal to the number of real roots of the 
given equation comprised between p and q. 

(308.) In order to simplify the demonstration of this theorem, 
we shall premise two Lemmas; and for convenience we shall 
call X the primitive function, and X,, X,,, X,,,, &c., auxiliary func- 
tions. 

Lemma 1. Of the series X, X,, X,,, &e., two consecutive functions 
cannot both vanish for the same value of x. 

From the method in which X,, X,,, &c., are obtained, we have 
the following equations : 


GENERAL THEORY OF EQUATIONS. 26: 


Xe eRe —X, (1). 
X, a X,, Q,, Tai > (2). 
X,, = XQ), ag ie (3). 


iT9 ce 74 
XKipg' = Be 1 — Ah (m — 1). 


Now suppose X,, = 0, and X,,, = 0; then, by equation (3), we 
have X,,,,= 0. Hence, since X,,,=0,:and X,;,,= 0; therefore, 
by equation (4), we must have X,— 0; and, proceeding in this 
manner, we shall find that X,, = 0, which is absurd, since it was 
shown, Art. 307, that this final remainder must be independent of 
x, and must therefore remain unchanged for every value of a. 


Lemma II. If one of the auxiliary functions vanishes for any par- 
ticular value of x, the two adjacent functions must have contrary signs 
for the same value of x. 

For, by equation (3), we have 


egy = Dt, 7, yr ai, : 


and if X,,,= 0, then X,, = — X,,,,; that is, X,, and X,,,, have con- 
trary signs. 


DEMONSTRATION OF THE THEOREM. 
(309.) Suppose all the real roots of the equations 
>. G— 0, Or 0, Xun 0, X= 0, &e., 


to be arranged in a series in the order of magnitude, beginning 
with the least. Let p be less than the least of these roots, and 
let it increase continually until it becomes equal to q, which we 
suppose to be greater than the greatest of these roots. Now so 
long as p is less than any of the roots, no change of signs will 
occur from the substitution of p for x in any of these functions, 
Art. 304; but when p arrives at a root of any of the auxiliary 
equations, its substitution for « reduces that polynomial to zero, 
and neither the preceding nor succeeding function can vanish 
for the same value of « (Lemma I.), and these two adjacent func- 
tions have contrary signs (Lemma II.). Hence the entire number 
of variations of sign is not affected by the vanishing of any of the 
auxiliary functions; for the three adjacent functions must re- 
duce to 


+,0,—, or —, 0, +. 


266 GENERAL THEORY OF EQUATIONS. 


Here is one variation, and there will also be one variation if we 
supply the place of the 0 with either + or —; thus, 


cise Ts Soy Oli 5 a rs 
hi. ee Tt ae 


Suppose, now, p to pass from a number very little smaller, to a 
number very little greater than a root of the primitive equation 


Re 0. 


the sign of X will be changed from + to —, or from — to +, Art. 
304. The signs of X and X, constitute a variation before the 
change, and a permanence after the change. Hence the change 
of sign of the function X occasions a loss of one variation of sign. 

Again, while p increases from a number very little smaller to a 
number very little greater than another root of X=0, a second 
variation will be changed into a permanence, and so on for the 
other roots of the primitive equation. 

Hence the number of variations Jost, when p increases from — 
a& to + o, must be equal to the whole number of real roots of the 
equation X = 0. 


(310.) EXAMPLES. 


Ex. 1. How many real roots has the equation 
x — 6a? + lle—_6=—01 
Here we have X = 32? — 122 + 11. 


Dividing 2° —6x°-+ 11x —6 by 32? — 122+ 11, as in the 
method for finding the greatest common divisor, Art. 250, we 
have for a remainder —2v+4. Hence, rejecting the factor 2, 
X,,=«x—2. Dividing X, by X,,, we have for a remainder — 1. 
Therefore, X,,,= + 1. 


Hence we have 
X =2°— 62°+ 11x —6. 
X, = 3a? — 122 + 11. 
X,, =e — 2. 
D. Gt a 6 


GENERAL THEORY OF EQUATIONS. 267 


Now, if we substitute for # in these functions, the numbers 
given below, the signs of the results will be as follows: 


Assumed Values of z. Resulting Signs. Variations. 
— on —+—+ giving 3 variations. 
0 pac Wea 2 ee 73 2 73 
+ 9 —+—+ 6 3 ¢ 
+1 Ch Ba A oo Ts 2 %3 
+1.1 rege ee 2 ele “ 9 « 
+1.9 Be nth Fay gli “« 9 “ 
+2, iP 20 ee Oe « 
+2.1 ——-+ + pa 1 os 
+-2.9 —++4 + Capt: 
+3 O+++ Gah OF inet 
+3.1 Seach tech OT « 
+ © ++++ CO Ome Le 


Here the three roots of this equation are seen to be 1, 2, 3, and 
no change of sign occurs by the substitution for 2 of any number 
less than 1; but when p exceeds 1, there is a change of sign in 
the original equation from — to +, by which one variation is 
lost. When p= 2, two of the functions disappear simultaneously, 
showing that 2 is aroot of the second derived function as well as 
of the original equation, and a second variation of sign 1s Jost. 
Also, when p becomes equal to 3, a third variation is lost; and 
there are no farther changes of sign arising from the substitution 
of any numbers between 3 and + o. 

There are ¢hree changes of sign of the primitive function, two 
of the first auxiliary function, and one of the second auxiliary 
function ; but no variation is lost by the change of sign of any of 
the aint functions; while every change of sign of the primi- 
tive function occasions a loss of one variation. 


Ex. 2. How many real roots has the equation 
v—52?+ 8r—_1=—01 


Here we find 


Reese Se as eet 
X,, =22 — 31 
X,,, =— 2295 


968 ’ GENERAL THEORY OF EQUATIONS. 


When x = — o, the signs are — + ——, giving 2 variations; 
r= +o, &s su hg ALES «“ 1 “ 
Hence this equation has but one real root, and consequently 
must have two imaginary roots. 
Ex. 3. How many real roots has the equation 
xt — Qe? — Tx? + 107+ 10=012 
Here we have 
X = at —22°—T7T2*+ 107+ 10. 
X, =4e> — 62? — 142+ 103 or 2x? — 3a? — Te +5. 
X,, = 1a? — 232 — 45. 
Xj, = 1522 — 305. 
Xj, = + 524785. 
When « =— o, the signs are + — + — +, giving 4 variations; 
©—=+ a, 66 ++ +44, 73 0 bs 
Hence the four roots of this equation are real. 
If we try different values for x, we shall find that 


When « = — 3, the signs are + — + — +, giving 4 variations; 


a a oe ne a: 
e=—1, “© —+t+—4, * 3 # 
L= 7M, . $f ——+, « Q Tt 
e—-+ 1, ts aE pee meee ey) rT 
mie Day ng My ro a A seh a aie 


r= + 3, i i ee a ae 
Hence this equation has two positive roots between 2 and 3; 
one negative root between 0 and —1; and one negative root be- 
tween — 2 and — 3. 
Ex. 4. How many real roots has the equation 
v—7rtiT7—02 
Ans. Three: viz., two between 1 and 2, and one between — 3 
and — 4. | 
Ex. 5. How many real roots has the equation 
Qet — 1327+ 102 —19=01 
Ans. Two: one positive and one negative. 
Ex. 6. How many real roots has the equation 
x’ —.362* + 72a°— 3744+ 72=01 
Ans. Three. 


SECTION XX. 


SOLUTION OF NUMERICAL EQUATIONS. 


(311.) We will first consider the method of finding the znéegral 
roots of an equation, and will begin with forming the equation 
whose roots are 2, 3, 4, and 5. This equation must consist of 
the factors 


(x — 2) (e —'8) (2 —4) (e —5)=0. 


If we perform the multiplication (which is most expeditiously 
done by the method of detached coefficients shown in Art. 64), 
we obtain the equation 


x* — 142° + 712? — 1542 4+ 120=—0. 


We know that this equation is divisible by x—5. Let us per- 
form the division by the method of detached coefficients shown 
in Art. 80. 


Ane Be pC D V a 
1— 14+ 71— 154+ 120 |A—5= divisor. 
1— 5 | 1— 9 + 26 — 24 = quotient. 
— 9+71 
— 9445 
+ 26 — 154 
+ 26 — 130 
9421-120 
— 24+ 120. 


Supplying the powers of x, we obtain for a quotient 


a8 — 99? + 260 — 24 = 0. 


270 SOLUTION OF NUMERICAL EQUATIONS, 


This operation may be still farther abridged, as follows: 

Represent the root 5 by a, and the coefficients of the given 
equation by A, B, C, D,..... Ne 

We first manltinte a oe A, and subtract the product from B; the 
remainder, — 9, we mulony by a, and subtract the product frekn 
C; the PAiaie® + 26, we Bmlticly again by a, and subtract 
from D; the remainder, — 24, we multiply by a, and, subtracting 
from V, nothing remains. If we take care to change the sign of 
ud, we may substitute addition for subtraction in the above state- 
ment; and if we set down only the successive remainders, the. 
work will be as follows: 


TN Sige Clap LEG ha 
1— 14+ 71 — 154+ 120] 5 
1— 9+26— 24, mi 


and the rule will be, 


Multiply A by a, and add the product to B; set down the sum, 
multiply it by a, and add the product to C3 set down the sum, multi- 
ply wt by a, and add the product to D, and so on. If the equation is 
exactly divisible, the final product will be equal to the last term V, 
taken with a contrary sign. 


The coefficients above obtained are the coefficients of a cubic 
equation whose roots are 2, 3,4. The equation may therefore 
be divided by «—4, and the operation will be as follows: 


1—9 + 26—24| 4 
Lee ae! 


These, again, are the coefficients of a quadratic equation whose 
roots are 2 and 3.. Dividing again by e—3, we have 


[eb B46 1S 
ey 


which are the coefficients of the binomial factor z— 2. 
These three operations of division may be exhibited together 
as follows: 


1— 14 + 71— 154+ 120] 5, first divisor. 
I— 9+ 26— 24 4, second divisor. 
1— 5+6 3, third divisor. 
1— 2. 


SOLUTION OF NUMERICAL EQUATIONS. pay Bi 


(312.) The method here explained will enable us to find all the 
integral roots of an equation. For this purpose, we make trial 
of different numbers in succession, all of which must be divisors 
of the last term of the equation. If any division leaves a re- 
mainder, we reject this divisor; if the division leaves no remain- 
der, the divisor employed is a root of the equation. Thus, by a 
few trials, all the integral roots may be easily found. 


Ex. 2. Find the seven roots of the equation 
x! + a°—14e°— 14a + 492° + 492°— 360 — 36 = 0. 


We take the coefficients separately, as in the last example, and 
try in succession all the divisors of 36, both positive and nega- 
tive, rejecting such as leave a remainder. The operation is as 
follows: 


14+ 1—14— 14+ 49 + 49— 36 — 36 1, first divisor. 


1+ 2— 12— 26 + 23+ 72 + 36 2, second divisor. 
14+4— 4—34—45—18 3, third divisor. 
1+-74+17+17+ 6 — 1, fourth divisor, 
1+6+4+ 114. 6 — 1, fifth divisor. 
1+5+ 6 — 2, sixth divisor. 
14+ 3 — 3, seventh divisor. 


Hence the seven roots are, 
yoke ge gate pale pation) Haag) 
Ex. 3. Find the six roots of the equation 
a + 5a°— 812*— 852° + 964x° + 7802— 1584 — 0. 


14+ 5—81— 8549644 780—1584| 1. 
1+ 6—75—160+ 804+ 1584 a 
1+ 10—35— 300— 396 hvn6 


Pesagheb1 4.66 - 2. 
1+ 14+ 33 ie 3) 
14ntt bucsipy? 


| 


The six roots, therefore, are, 
1, 4, 6, —2, —3, — 11. 
Ex. 4, Find the five roots of the equation 
a + 6a'— 102*°— 1122? —207x— 110=—0. 


272 SOLUTION OF NUMERICAL EQUATIONS. - 


1+ 6— 10—112—207—110 | —1. 


1+5—15— 97—110 — 2. 
1+ 3—21— 55 pres 
1—2—11 


Three of the roots, therefore, are 
— 1, —2,.—5. 


The two remaining roots may be found by the ordinary meth- 
od of quadratic equations. Supplying the letters to the last co- 
efficients, we have 


av’ —2r—11—0. 
Hence z-—1tv 12. 


Ex. 5. Find the four roots of the equation 
x* + 22° — To? — 82 + 12—0. 
Ex. 6. Find the four roots of the equation 
x*— 55x* — 3027 + 504=0. 


Ex. 7. Find all the roots of the equation 


v'— 252° + 602 — 36 =0. Lo JLo eae 


Ex. 8. Find all the roots of the equation 
a + Bat + a? — 16x? — 20x — 16 = 0. 


HORNER’S METHOD. 


(313.) The preceding method furnishes the roots of an equa- 
tion only when they are expressed by whole numbers. When the 
roots are incommensurable, we employ the following method, 
which is substantially the same as published by Horner in 1819. 

The Theorem of Sturm, together with Art. 304, enables us to 
find the zntegral part of any real root of the equation proposed. 
We then transform the equation into another having its roots 
less than those of the preceding by the number just found, Art. 


302. We discover again, by Art. 304, the first figure of the root 


of this equation, which will be the first decimal figure of the root 
of the original equation. Again, we transform the last equation 
into another having its roots less than those of the preceding by 
this decimal figure. We thus’discover the second decimal figure 


SOLUTION OF NUMERICAL EQUATIONS. 273 


of the root; and proceeding in this manner from one transfor- 
mation to another, we are enabled to discover the successive fig- 
ures of the root, and may carry the approximation to any degree 
of accuracy required. 


Example 1. Find the roots of the cubic equation 
xv’ + 32° + 52— 178. 


The several functions given by Sturm’s Theorem are, 


X =f +327 + 5r—I1'8,. 
X, =32? +467 +5, 
X,, =— 4a + 539, 
X /;, = — 884579. 
Now for « = + o, the signs are, + + ——, giving 1 variation. 
v= —o, és wes ff a3 9 rT; 


Therefore two values of x are imaginary, and the equation has 
but one real root. 

We readily find, by the method of Art. 310, that this root lies 
between 4 and 5. The first figure of the root, therefore, is 4. 
To ascertain the second figure, we transform the given equation 
into another in which the value of x is diminished by 4, which 
is done by substituting for v7, y-++4. We thus obtain 


y+ 15y? + T7y = 46. 
The first figure of the root of this equation, according to Art. 
304, is .5. Now transform the last equation into another in 


which the value of y is diminished by .5, which is done by sub- 
sfituting for y, z+ .5. We thus obtain 


2+ 16.52? + 92.752 = 3.625. 


The first figure of the root of this equation is .03. We must 
now transform this equation into another in which the value of 
z is diminished by .03, which is done by substituting for z, 
y+ .03. We thus obtain | 


v + 16.590" + 93.7427 = .827623. 


The first figure of the root of this equation is .008. 

In order to find the next figure, we must transform the last 
equation into another in which the value of v is diminished by 
.008, and so on. 

S 


274: SOLUTION OF NUMERICAL EQUATIONS. 


(314.) This method would be very laborious if we were obli- 
ved to deduce the successive equations from each other by the 
ordinary method of substitution; but they may all be derived 
from each other by a very simple law. Thus, let 


Az? +Bx’+Cx=D (1) 
be any cubic equation; and let the first figure of its root be de- 
noted by a, the second by a’, the third by a’, and so on. 
If we substitute a for x in equation (1), we shall have 
Ad + Ba’? + Ca =D. 
D 


Whence a —C+ Ba} Aa (2). 


If we put y for the sum of all the figures of the root except 
the first, we shall have x =a+y; and eu nee this value for x 
in equation (1), we obtain 

Ad + 3Aa’y + 3Aay’ + Ay’ 
+ Ba? + 2Bay + By | De 
+ Ca +Cy 
or, arranging according to the powers of y, 


Ay’ + (B+ 3Aa)y’?+(C + 2Ba + 3Aa’)y = D—Ca— Ba’— Aa” (3). 
Let us put B’ for the coefficient of y°*, C’ for the coefficient of 
y, and D’ for the right member of the equation, and we have 
Ay + By?+Cy=D’ (4). 


This equation is of the same form as equation (1) 3 and pro- 
ceeding in the same manner, we shall find 


D' 
Y=OT Ba paa® ©) 
where a’ is the first figure of the root of equation (4), or the sec- 
ond figure of the root of equation (1). 


Putting z for the sum of all the remaining figures, we have 
y= a’+2z; and substituting this value in equation (4), we shall 
obtain a new equation of the same form, which may be written 


Az ve Bz? a C’ z—D” (6) ; 


SOLUTION OF NUMERICAL EQUATIONS. 275 


and in the same manner we might proceed with the remaining 
figures. 

Equation (2) furnishes the value of the first figure of the root; 
equation (5) the second figure, and similar equations would fur 
nish the remaining figures. Each of these expressions involves 
the unknown quantity which is sought, and might therefore ap- 
pear to be useless in practice. When, however, the root has 
been already found to several decimal places, the value of the 
terms Ba and Aa’ will be very small compared with C, and a will 


D 
be very nearly equal to CG C, then, may in all cases be employ- 


ed as an approximate divisor, which will probably furnish a new 
figure of the root. Thus, in the above example, all the figures 
of the root after the first, are found by division. 
LT lis wy Wf Spa 
a O20 seta aatUds 
827 = 93.74 = .008. 


If we multiply the first coefficient A by a, the first figure of 


the root, and add the product to the second coefficient, we shall 
have 


B+ Aa (7). 
If we multiply this expression by a, and add the product to the 
third coefficient, we shall have 
C+Ba+Aa’® (8). 
If we multiply this expression by a, and subtract the product 
from D, we shall have 
D—Ca— Ba?— Aa’, 
which is the quantity represented by D’ in equation (4). 
Again, multiplying the first coefficient by a, and adding the . 
product to expression (7), we obtain 
B+2Aa (9). 


Multiplying this expression by a, and adding the product to 
expression (8), we have 


276 SOLUTION OF NUMERICAL EQUATIONS, 


! C + 2Ba + 3Aa’, 
which ts the coefficient of y in equation (4). 


Again, multiplying the first coefficient by a, and adding the 
product to expression (9), we have 


B + 8Aa, 
which is the coefficient of y’ in equation (4). 


We have thus obtained the coefficients of the first transformed 
equation ; and by operating in the same manner upon these coef- 
ficients, we shall obtain the coefficients of the second transform- 
ed equation, and so on; and the successive figures of the root are 
found by dividing D by C, D’ by C’, D” by C”, and so on. 

(315.) The preceding method is summed up in the following 


RULE. 


Represent the coefficients of the different terms by A, B, C, and the 
right-hand member of the equation by D. Having found a, the first 
figure of the root, multiply A by a, and add the product to B. Mul- 
tiply this sum by a, and add the product to C. Multiply this last sum 
by a, and subtract the product from D ; the remainder will be the First 
DIVIDEND. 

Again, multiply A by a, and add the product to the last number un- 
der B. Multiply this sum by a, and add the product to the last num- 
ber under C3; this last sum will be the FIRST DIVISOR. 

Again, multiply A by a, and add the product to the last number un- 
der B. 

Find the second figure of the root by dividing the first dividend by 
the first divisor, and proceed with this second figure Perse as was 
done with the first figure. 

The second figure of the root obtained by division, will fre- 
quently furnish a result too large to be subtracted from the re- 
mainder D’, in which case we must assume a different figure. 
After the second figure of the root has been obtained, there will 
seldom be any farther uncertainty of this kind. 

The operation for finding a root of the equation 


ot 3x? + 5a — 178 


will then proceed as follows: 


SOLUTION OF NUMERICAL EQUATIONS. Oia 








Ae 2tB Cc D a 
1 +3 +5 =178 (4.5388 = 2. 
4 38 132 
i 33 46 — 1st dividend. 
Pee 4.4 42.375 
rT 77 = Ast divisor. 3.625 — 2d dividend. 
4, 1.15 Tw oe oe 
15.5 84.75 .827623 — 3d dividend. 
5 8.00 .75 1003872 
16.0 92.75 — 2d divisor. .076619128 = 4th dividend. 
‘5 A959 
16.53 93.2459 
3 4968 
16.56 93.7427 = 3d divisor. 
3 132784 
16.598 93.875484 
8 .132848 








16.606 94.008332 = 4th divisor. 
Having found one root, we may depress the equation 
a + 32? + 5¢2—178 = 0 
to a quadratic, by dividing it by « — 4.5388. We thus obtain 
x + 7.53882 + 39.2173 = 0, 


where = is evidently imaginary, because g is negative and great- 
2 


er than. See Art. 195. 


After thus obtaining the root to five or six decimal places, sev- 
eral more figures will be correctly obtained by simply dividing the 
last dividend by the last divisor. 


Example 2. Find the roots of the equation 
x + 112? — 1022 = — 181. 


The first figure of one of the roots we readily find to be 3. 
We then proceed, according to the Rule, to obtain the root to 
four decimal places, after which two more will be obtained cor- 
rectly by division. 


278 SOLUTION OF NUMERICAL EQUATIONS. 


A B C D a 
tue a) — 103 = —181 (3.21312 =2. 
3 42 — 180 
“14 60 — 1 = 1st dividend. 
3 51 — .992 
17 = —9 = Ist divisor, | — .008 — 2d dividend. 
3 4.04 — 006739 
20.2 —4.96 001261 = 3d dividend. 
2 4.08 — .001217403 
20.4 — 0.88 = 2d divisor. — .000043597 — 4th dividend 
2 2061 
20.61 —.6739 
1 2062 
20.62 — 4677 — 3d divisor. 
1.061899 
~ 20.633 — .405801 
3 061908 





20.636 — .343893 — 4th divisor. 


The two remaining roots may be found in the same way, or by 
depressing the original equation to a quadratic. Those roots are, 


3.22952 
— 17.44.265. 


When a power of z is wanting in the proposed equation, we 
must supply its place with a cipher. | 
Ex. 3. Find all the roots of the cubic equation 


x —iz = —7 


The work of the following example is exhibited in an abbrevia- 
ted form. Thus, when we multiply A by a, and add the product 
to B, we set down simply this result. We do the same in the 
next column, thus dispensing with half the number of lines em- 
ployed in the preceding example. Moreover, we may omit the 
ciphers on the left of the successive dividends, if we pay proper 
attention to the local value of the figures. Thus, it will be seen 
that in the operation for finding each successive figure of the 


SOLUTION OF NUMERICAL EQUATIONS. . 279 


root, the decimals under B increase one place, those under C in- 
crease ¢wo places, and those under D increase three places. 


1 +0 —7 ——7 (1.356895867 = a. 
1 betty aos 
2 —_4,— 1st div’r. —1-= 1st dividend. 
Pd method mache | 
3.6 1.93 =2d div’r, —-97=2d dividend. 
3.95  —1.7325 86625 
4.00 —1.5325—3ddiv’r. 10375 — 3d dividend. 
4.056  —1.508164 9048984 
4.062 .—1.483792 =4thdiv’r. 1326016 = 4th dividend 
4.0688 —1.48053696 1184429568 
4.0696 — 147728128 = dthdiv’r. 141586432 — 5th div’d. 
4.07049 —1.4°769149359 132922344231 


4.07058 — 1.4:765485837 = 6thdiv’r. 8664087769 = 6thdiv’d. 


Having proceeded thus far, four more figures of the root, 5867, 
are found by dividing the sixth dividend by the sixth divisor. 


We may find the two remaining roots by the same process 5 
or, after having obtained one root, we may depress the equa- 
tion 


V—zt+T=—0 -s 


to a quadratic equation, by dividing by « — 1.356895867, and we 
shall obtain 


a + 1.356895867x — 5.158833606 = 0. 
Solving this equation, we obtain 


x= — .678447933 + JV 5.619125204 
= — 3,048917338, or + 1.692021471, 


the two remaining roots. 


Ex. 4. Find a-root of the equation 


22° + 3x? = 850. 


31 

45.10 
4.5.20 
45.3004 
45.3008 
45.30130 
45.30140 


SOLUTION OF NUMERICAL EQUATIONS. 


0 —850 (7.0502562208 
119 833 
336—1st divisor. | 17=1st dividend. 
338.2550 16.912750 
340.5150—2d divisor. 87250—2d dividend. 
34.0.524.06008 68104812016 
34.0.53312024—8d div. 19145187984—3d div’d. 
34.0.5353853050 1'7026769265250 


34.0.5376503750 =4th div. 2118418718750 —4th div. 


Dividing the fourth dividend by the fourth divisor, we obtain 
the figures 62208, which make the root correct to the tenth deci- 


mal place. 


The two remaining values of 2 may be easily shown to be ima 


ginary. 


When a negative root is to be found, we change the signs of 
the alternate terms of the equation, Art. 299, and proceed as for 
a positive root. 


Ex. 5. Find a root of the equation 


dx* — 62? + 32 = — 85. 


Changing the signs of the alternate terms, it becomes 


5 +6 
16 
26 
36.5 
37.0 
37.80 
38.10 
38.405 
38.410 
38.4165 
38.4180 


5a? + 62° + 382 — + 85. 


te +85 (2.16139. 
35 10 
87— Ist divisor. 15 = Ist dividend. 
90.65 9.065 
94.35 —2d divisor. 5.935 — 2d dividend. 
96.6180 5.797080 
98.9040 = 3d divisor. 137920 — 3d dividend. 
98.94.2405 9894.24.05 
98.980815 =4th div’r. 38977595 — 4th div’d 
98.99232995 29697701985 





99.00386535 —5thdiv’r. 9279893015 = 5th div’d. 


SOLUTION OF NUMERICAL EQUATIONS. esl 


Hence one root of the equation 
5a* — 62° + 3x = — 85 
is — 2.16139. 


The same method is applicable to the extraction of the cube 
root of numbers. 

Ex. 6. Let it be required to extract the cube root of 9; m 
other words, it is required to find a root of the equation 


= 9. 
120 0 9 (2.0800838. ° 
2 4 8 
4, 12= 1st divisor. 1=1st dividend. 
6.08 12.4864 998912 
6.16 12.9792—2d divisor. 1088 = 2d dividend. 
6.24008  12.9796992064 1038375936512 
6.24016 12.9801984192—3ddiv. 49624063488 — 3d div. 
6.240243 12.980217139929 — - 3894.0651419787 


6.240246  12.980235860667—4th d. 10683412068213—4th d. 


Ex. 7. Find one root of the equation 
x? + 2? = 500. 
Ans. @ = 7.617280. 
Ex. 8. Find one root of the equation 
a + 2* + 2 = 100. 
Ans. v = 4.264430. 
Ex. 9. Find one root of the equation 
Q2° + 32? — 4a = 10. 
Ans. 2 = 1.624819. 
Ex. 10. Extract the cube root of 48228544. 


é 


Ans. 364. 


Ex. 11..There are two numbers whose difference is 2, and 
whose product, multiplied by their sum, makes 120. What are 
those numbers 2? 


aon 
o 


5 


282 SOLUTION OF NUMERICAL EQUATIONS. 


Ex. 12. It is required to find two numbers whose difference is 
6, and such that their sum, multiplied by the difference of their 
cubes, may produce 5040. 


Ex. 13. Required two numbers whose difference is 5, and such 
that, if the less be multiplied by the square root of the giving! 
the Arodiet may be 12. 


Ex. 14. There are two numbers whose difference is 4; and the 
product of this difference, by the sum of their cubes, is 3416. 
What are the numbers 2 


Ex. 15. Several persons form a partnership, and establish a 
certain capital, to which each contributes ten times as many dol- 
lars as there are persons in company. They gain 6 plus the 
number of partners per cent., and the whole profit is $392. 
How many partners were there 2 


Ex. 16. A company of merchants have a common stock of 
$4775, and each contributes to it twenty-five times as many dol- 
lars as there are partners, with which they gain as much per cent 
as there are partners. Now on dividing the profit, it is found, 
after each has received six times as many dollars as there are 
persons in the company, that there still remains $126. Required 
the number of merchants 2 


Ans. 7, 8, or 9. 


EQUATIONS OF THE FOURTH AND HIGHER DEGREES. 


(316.) The method already explained for cubic equations is ap- 
plicable to equations of every degree. For the fourth degree, we 
shall have one more column of products, but the operations are 
all conducted in the same manner, as will be seen from the fol- 
lowing example. 


Ex. 1. Find the four roots of the equation 
at — 8a? + 14a? + 4a — 8. 


By Sturm’s Theorem, we find that these roots are all real; 
three positive, and one negative. 


We then proceed as follows: 


: 8 


SOLUTION OF NUMERICAL EQUATIONS. 283 

1 —8 + 14 + 4 =8 (5.2360679. 

—3 — 1 — i —)5 “a 

+2 +9 +44— 1st divisor. 13—1st dividend. 

7 4.4, 53.288 10.6576 

1%,2 4:6.4:4 63.072=—2d div’r. 2.3424—2d dividend. 

12.4 48.92 64.626747 1.93880241 

126 5144 66.193068—3dd. .40359759=3d div’d. 

12.83 51.8249 66.509117736 | .399054;706416 

12.86 52.2107 66.825633024—4thd. 4542883584—4th d. 

12.89 52.5974 4 


12.926 52.674956 
12.932 52.752548 


and by division we obtain the four figures 0679. 
x= 2.7320508, 
The other three roots are 2 == .7639320, 
x = — .7320508. 
Ex. 2. Find the roots of the equation 


xv + Qa* + 32° + 4a? + 5a = 20. 


Here we have, according to Sturm’s Theorem, 


X = ww + 2* + 32° + 40° + 5x — 20, 
Xx, = 5at + 82? + Ox + Se 4+ 5, 
X,, = — Te? — 1a? — 42x + 255, 
> ee ee ee 
DOF ie 
When x = + o, the signs are + + — — —, one variation. 
i oe " —+++4+-,two “ 


Hence the equation has one real and four imaginary roots. 


We then proceed as follows: 


* 


SOLUTION OF NUMERICAL EQUATIONS. 


284 


“AIP WIP = GL89F0EL86|9ECE 

CBLESLEGLG|98VBB" 

‘PUEPIAIP PE = 89GES|ETBGT' e 
GEOPPIGOOLS’ 

‘PUSPIAIP PZ = GE8VI'T 

ILILS’S 

‘puopralp Is] = G 

GT 
“68LGZE'L) Oo+ 


“AIP UF = CBI E0GES GOGL' PF 
CZ90GLPP TELG PP 

‘LOSIAIP PE = 0806 /99CE PF 
910G LZ0G'SF 

“LOSIAIP PZ = G8C9'GP 
TLIL'8€ 

*LOSIAIP IST = CE 


cT 


G+ 


OOGO69ILEV ED 
CZLO6L| POE EP 
O80|G8l'EP 
GE0/S69 GP 
800'TTGGP 
O€L TP 
TIT 6E 
TLULE 
cE 
0G 
OT 

P+ 


GLOO|8SS OL/9IL 
GB0G|TS VG COOL 


OF/0G' VG 
£O|GE'VG 
GL|0G'VG 
70|SO"'PG 
06°E% 
91'E% 
EPCS 
IL'T@ 
1G 

ct 


Sigh 
9h 
PIL 
GIGh 
VL 
oF 
Gh 
TL 

9 


SOLUTION OF NUMERICAL EQUATIONS. 285 


Dividing the fourth dividend by the fourth divisor, we obtain 
the figures 789. 

When we wish to obtain a root correct to a limited number of 
places, we may save much of the labor of the operation by cut- 
ting off all figures beyond a certain decimal. Thus if, in the ex- 
ample above, we cut off all beyond five decimal places in the suc- 
cessive dividends, and all beyond four decimal places in the divi- 
sors, it will not affect the first six decimal places in the root. 


Ex. 3. Find the roots of the equation 
xt — 122? + 1227 — 3. 
f 2 — + 2.858083, 


a 2 = + 443277, 


@ = — 3.907378. 
Ex. 4. Find the roots of the equation 
xt — 82° + 14a? + 42 = 8. 
xv = + 5.236068, 
EE arto 
x= — .732050. 
x. 5. Find the roots of the equation 


vc’ — 102° + 62 = — 1. 


eet 
i*. 


© = — 3.065315, 
| e=— .691576, 
Ans. ¥ «= — .175675, 


— + .879508, 
| x = + 3.053058. 


8 


Ex. 6. Find one root of the equation 
. xe + 3at + 22? — 32? — 2n — 2. 
7 Ans. 7 = 1.059109. 


Ex. 7. Required the fourth root of 18339659776. 
Ans. 368. 


Ex. 8. Required the fifth root of 26287667869473. 
Ans. 483, 


286 SOLUTION OF NUMERICAL EQUATIONS. 


RESOLUTION OF EQUATIONS BY APPROXIMATION. 


(317.) The method of Horner for finding the incommensurable 
roots of a numerical equation is generally better than any other; 
nevertheless, the method by approximation may sometimes be 
preferred. We shall explain the method of Newton, and that of 
Double Position. 


METHOD OF NEWTON. 


This method supposes that we have already determined nearly 
the value of one root; that we know, for example, that such a 
value exceeds 4, and that it rs less than 5. In this case, if we 
suppose the exact value —4 + y, we are certain that y expresses 
a proper fraction. Now as y is less than unity, the square of y, 
its cube, and, in general, all its higher powers, will be much less 
with respect to unity ; and for this reason, since we only require 
an approximation, they may be neglected in the calculation. 
When we have nearly determined the fraction y, we shall know 
more exactly the root 4+ y; from which we proceed to deter- 
mine a new value still more exact, and we may continue the ap- 
proximation as far as we please. 

We will illustrate this method by an easy example, requiring 
by approximation the root of the equation 


ro alls 
Here we perceive that z is greater than 4, and less than 5. It 
we suppose z — 4:+ y, we shall have 
x = 16 + 8y+ 7? = 20. 
But, as y* must be quite small, we shall neglect it, and we have 
16 + 8y = 20, or 8y = 4. 
Whence y = .5, and « = 4.5, which already approaches near 
the true root. If we now suppose 2 = 4.5 + z, we are sure that 


z expresses a fraction much smaller than y, and that we may 
neglect z? with greater propriety. We have, therefore, 


2 = 20.25 + 92 = 20, or 92 = —.25. 


Consequently, z= — 0278. 
Therefore, v.= 4.5 — .0278 = 4.4722. 


A See 


“4 


st 
* 


ee? a 


SOLUTION OF NUMERICAL EQUATIONS. 287 


If we wish to approximate still nearer to the true value, we 
must make a — 4.4722 + v, and we should have 


a? — 20,00057284 + 8.94440 = 20. 
So that 8.94449 — — .00057284. 
Whence v = — .0000640. 
Therefore, « — 4.4722 — .0000640 — 44721360, 


a value which.is correct to the last decimal place. 
(318.) In order to generalize this method, suppose the given 
equation to be 


eo = D, 


and that we have found z to be greater than a, but less than 
a+1. If we make x =a-+ y, y must be less than unity, and y’ 
may be neglected, as a very small quantity ; so that we shall 
have 

ea + 2ay'= D, 


or ay =D—at; and y = > 


DP —a’? Bat, D, 


Consequently, #w=a-+ og 





Now, if a was near the true value of x, this new expression 
a+D . CSS 
ae will approach nearer; and, by substituting it for a, we 
shall obtain a new value, which may again be employed in order 
to approximate still nearer, and the same operation may be con- 
tinued as long as we please. 

This method of approximation may be employed in finding the 
roots of all equations. 


Suppose we have the general equation of the third degree, 
Ax* + Ba’? + Cx = D, 


in which a is very nearly the value of one of the roots. Let us 
make «= a-+ y; and, since y is less than unity, neglecting the 
powers above the first degree, we shall have 

2 = a* + 2ay, 

x = a+ 3a’y, 


288 SOLUTION OF NUMERICAL EQUATIONS. 


whence we have the equation 
Aad’ + 3Aa’y + Ba? + 2Bay + Ca + Cy = D. 


Wh _ D—Ca— Ba’ — Ad} 
se I= C4 2Ba + 3Ae 


The numerator of this fraction is the quantity represented by 
D’ on page 274; that is, it is the dividend in Horner’s method, 
and the denominator is the divisor; and we have already seen 
that this quotient furnishes one or more figures of the root. 


EXAMPLES. 
Ex. 1. For an application of this method, take the equation 
x + Q2° + 3x7 = 50. 
Here A = 1;'B = 2) 'C ='8,"and D = 150: 


pee aden! 2 
Therefore, Vis SSS 


We see that x is nearly equal to 3. If we substitute 3 for a, 
we shall have 
y = a7 
Whence x = 2.9 nearly. 


And if we substitute this new value instead of a, we shall find — 
another still more exact. 
Ex. 2. For another example, take the equation 


x —6x = 10. 
If we make « = a + y, we shall have 
a’ — a’ + 5aty, 


neglecting the higher powers of y, and, consequently, 


a’ + 5a'y — 6a — by = 10. 


10 + 6a — a? 
Therefore, ee. 
Assume a = 2, and we obtain 
*:104+.12— 382 » '5 


fre ay RR Py 


i 


SOLUTION OF NUMERICAL EQUATIONS. 989 


Hence x = 1.86 nearly. 
If we assume a = 1.86, we have 


10 + 11.16 — 22.262 | 


mt 59.844 — 6 Tor mer 
Hence z = 1.839 nearly. - 
If we assume a = 1.839, we shall have 
4, AO LI Oo4e—— 21.033352 — 00001266. 


57.18694 — 6 
Therefore, z = 1.83901266. 


Ex. 3. Given x — 9x = 10, to find one value of x by approxi- 
mation. 


Ans. x = 3.4494897, 


Ex. 4. Given 2 + 92? + 4x2 = 80, to find one value of x by ap- 
proximation. 


Ans. x = 2.4'°721359. 


METHOD OF DOUBLE POSITION. 


(319.) Another method of finding the roots of an equation is 


by the rule of Double Position. 


Substitute in the given equation two numbers as near the true 
root as possible, and observe the separate results, Then state the 
following proportion : } 


Als the difference of these results, 

Is to the difference of the two assumed numbers, 

So is the error of either result, 

To the correction required in the corresponding assumed number. 


This being added to the number when too small, or subtracted 
from it when too great, will give the true root nearly. The num- 
ber thus found, combined with any other that may be supposed 
to approach still nearer to the true root, may be assumed for an- 
other operation, which may be repeated till the root is determined 
to any degree of accuracy required. | 


290 SOLUTION OF NUMERICAL EQUATIONS. 


EXAMPLES. 


Ex. 1. Given a + x? + x = 100, to find an approximate vatue 
of 2. 

Having ascertained, by a few trials, that ~ is more than 4, and 
less than 5, let us substitute these two numbers in the given 
equation, and calculate the results. 


iii ao 

By the first sup- ar _1¢ By the second sup- a hee 
position, get) position, oF ee RD 
Result, 84: Result, 155 


Then 155 — 84:5 —4:: 100 — 84: .22. 


Therefore, 4 + .22, or 4.22, approximates nearly to the true 
root. 

If, now, 4.2 and 4.3 be taken as the assumed numbers, and sub- 
stituted in the given equation, we shall obtain the value of 
x = 4.264 nearly. 

Again, assuming 4.264 and 4.265, and roeeecinen in the same 
manner, we shall find « = 4.2644299 very nearly. 

This rule is founded on the supposition that the differences m 
the results are proportioned to the differences in the assumed 
numbers. This supposition is not strictly correct ; but if we em- 
ploy numbers near the true values, the error commonly is not very 
great, and it becomes less and less the farther we carry the ap- 
proximation. 


Ex. 2. Given a + 2x’ — 23x — 70 = 0, to find one value of @. 
i Ans. x = 5.13450. 


Ex. 3. Given 2° — 15x? + 632 — 50 = 0, to find one value of x 
Ans. 7 = 1.028039. 


Ex. 4. Given at — 32° — Tbe = 10000, to find one value of z. 
Ans. « = 10.2615. 
Ex. 5. Given a + 2z* + 32° + 4x? + 5x2 = 54821, to find one 
value of a. 
Ans. x = 8.4144. 


SOLUTION OF NUMERICAL EQUATIONS. 291 


(320.) We will conclude this Section by finding the different 
roots of unity. 

Ex. 1. Find the two roots of the equation x’ = 1, or the square 
roots of unity. 

Extracting the square root, we find 


c—-+ 1, or—1. 
Ex. 2. Find the three roots of the equation, x* = 1, or the cube 


roots of unity. 


Since one root of this equation is z= 1, the equation 2°—1=0 
must be divisible by e— 1; and dividing, we obtain 


“e+e+ti=0;3° 


a eek (Piak 
whence e=—$44¥—3, r= —* 


Hence the required roots are, 


ee ite! 3 1 8 
+1 SS, SS 


which are the cube roots of unity. 


These results may be easily verified. We have seen, on page 
282, that the cube of —1+ / —3 is 8, which, divided by 8, 
the cube of the denominator, gives +1, as required. 


Ex. 3. Find the four roots of the equation z* = 1, or the fourth 
roots of unity. 


The square root of this equation is 
“2—+1,or=—1. 
Hence the required roots are, 
+1,—1,+¥—1, —V—1. 


Ex. 4. Find the five roots of the equation x = 1. 


Since one root of this equation is « = 1, the equation 2° — 1 
‘must be divisible by e— 13 and dividing, we obtain 


etet+eft+etl1l=4o. 


299 SOLUTION OF NUMERICAL EQUATIONS. 


Dividing, again, by 2’, we have 


Bie | 


1 
Now put z=2+ ~ 
1 
Whence 2a=e74+20+ > 


which, being substituted in equation (1), gives 
2+2—1—0. 


This equation, solved by the usual method, gives 


1 
bq tH == : 
ae 
are, 
z Zz — 4 z 24 
2 =244/ ii , and x => — tt 





in which the value of z being substituted, gives 


PA oe a OVP keen eee ER TPN GAY | 


or oe )/ 5 ed ee Oe 
Hence the five fifth roots of unity are, 
1 


Bape habe ree) Seo a 
tL 5 eee 2 o/ Bal, 
—i[v5+1—v—0+2V51. 
—i[v54+14+V—10F2V61. 


Ex. 5. Find the six roots of the equation 2° = 1. 


SOLUTION OF NUMERICAL EQUATIONS, 293 


These are found by taking the square roots of the cube roots. 
Hence we have, 


fe Ly ferads 4447 —3, —}4}—V7—3. 


Thus we see that unity has ¢wo square roots, ¢hree cube roots, 
four fourth roots, five fifth roots, sz sixth roots, and, generally, 
n nth roots; or the mth root of unity admits of » different alge- 
braic values. As, however, most of these roots are imaginary, 
they cannot be found by Horner’s Method. 


an 


SECTION XXL. ¥- aun 6 


LOGARITHMES. 


(321.) In a system of logarithms, all numbers are considered 
as the powers of some one number, arbitrarily assumed, which is 
called the base of the system; and the exponent of that power of the 
base which is equal to any given number ts called the logarithm of that 
number. 

Thus, if a be the base of a system of logarithms, and @ SN; 


ig to-whichthe_base (a) must-be meee to- equal N. 

If a* = N’, then 3 is the logarithm of N’ for the same reason ; 

And if a* = N”, then @ is called the logarithm of N” in the 
system whose base is a. 

The base of the common system of logarithms (called, from 
their inventor, Briggs’ Logarithms) is the number 10. Hence 
all numbers are to be regarded as powers of 10. Thus, since 


10° = 4 0 is the logarithm of 1 in Briggs’ system. 
10' = 2G, 1 cs 10 A 
10° =.108, ae . 100 BS 
10? = 1000, 3 eS 1000 et 
10* = 10000, 4 # 10000 i: 
&c., &ec., &c. 


From this it appears that, in Briggs’ system, the logarithm of 
every number between 1 and 10 is some number between 0 and 1, 
7. @, is a proper fraction. The logarithm of every number be- 
tween 10 and 100 is some number between 1 and 2, 7. e., is 1 plus 
a fraction. The logarithm of every number between 100 and 


ee 
then 2 is the logarithm of N; that is, 2 is the exponent of the //77/ 


/ 


i her 


LOGARITHMS. 295 


1000 is some number between 2 and 3, 2. ¢., is 2 plus a fraction, 
and so on. 

(322.) The preceding principles may be extended to fractions 
by means of negative exponents. Thus, 

fox oF 107'> 0.15 therefore, —1 is the logarithm of .1 


in Briggs’ system. 


zis or 10°=0.01; 6 — 2 ue 01 
toss or 10°=0.001; a —3 ff 001 
rotss or 10°*= 0.0001 ; 66 — 4 it 0001 


Hence it appears that the logarithm of every number between 
1 and .1 is some number between 0 and — 1, or may be repre- 
sented by —1 + a fraction; the logarithm of every number be- 
tween .1 and .01 is some number between — 1 and — 2, or may 
be represented by —2-+ a fraction; the logarithm of every 
number between .01 and .001 is some number between —2 and 
— 3, or is equal to —3-+ a fraction, and so on. 

The logarithms of most numbers, therefore, consist of an inte- 
ger and a fraction. The integral part is called the characteristic 
or index, and may always be known from the following 


RULE. 


The index of the logarithm of any number greater than unity, is one - . 
less than the number of integral figures in the given number. 

Thus, the logarithm of 297 is 2+ a fraction; that is, the cndex 
of the Jogarithm of 297 is 2, which is one less than the number 
of integral figures. The index of the logarithm of 5673 is 3; of 
73254, is 4, &e. 

The index of the logarithm of a decimal fraction ts a negative num- 
ber, and ts equal to the number of places by which tts first significant 
jigure is removed from the place of units. 

Thus, the logarithm of .0046 is —3-+ a fraction; that is, the 
index of the logarithm is —3, the first significant figure, 4, being 
removed three places from units. 


B00.” LOGARITHMS. 


GENERAL PROPERTIES OF LOGARITHMS. 


(323.) Let N and N’ be any two numbers, x and 2’ their re- 
spective logarithms, a the base of the system. Then, by the def- 
inition, Art. 321, 

IN“ a? (1); 
N’ = a” (2). 
Multiplying together equations (1) and (2), we obtain 


NN’ = ata?” 
meee qzto* 


Therefore, according to the definition of logarithms, x + 2’ is 
the logarithm of NN’, since 2 + 2” is the index of that power of 
the base a which is equal to NN’; that is to say, 

The logarithm of the product of two or more factors is equal to the 
sum of the logarithms of those factors. 

Hence we see that if it is required to multiply two or more 
numbers by each other, we have only to add their logarithms; the 
sum will be the logarithm of their product. We then look in the 
Table for the number answering to that logarithm, in order to 
obtain the required product. 


EXAMPLES. 


Ex. 1. Find the product of 8 and 9 by means of logarithms. 


In Art. 334, the logarithm of 8 is given 0.903090 
i ha 0.954.243 
The sum of these two logarithms ts 1.857333 


which, according to the same Table, is seen to be the logarithm 
of 72. 


Ex. 2. Find the continued product of 2,5, and 14 by means of 
logarithms. 


Ex. 3. Find the continued product of 2, 3, 4, and 5 by means 
of logarithms. 


LOGARITHMS. a | 
[f, instead of multiplying, we divide equation (1) by equation 
(2), we shall obtain | 
N a” a 
ee as Cb ° 
Neer a° 


Therefore, according to the definition, e—' is the logarithm 





oe 
of —-, since «—z2' is the index of that power of the base a 


N’ 
N 
w that is to say, 

The logarithm of a fraction, or of the quotient of one number divi- 
ded by another, is equal to the logarithm of the numerator, minus the 
logarithm of the denominator. 

Hence we see that if we wish to divide one number by anoth- 
er, we have only to subtract the logarithm of the divisor. from 
that of the dividend; the difference will be the logarithm of their 
quotient. 


_ which is equal to 


EXAMPLES. 


Ex. 1. It is required to divide 108 by 12 by means of loga- 
rithms. 


The logarithm of 108 is . 2.033424 
12 1.079181 
The difference is 0.954.243, 


which is the logarithm corresponding to the number 9. 

Ex. 2. Divide 133 by 7 by means of logarithms. 

Ex. 3. Divide 136 by 17 by means of logarithms. 

Ex. 4. Divide 135 by 15 by means of logarithms. 

The preceding examples are designed to illustrate the proper- 
ties of logarithms. In order to exhibit fully their wezlity in com- 
putation, it would be necessary to employ larger numbers; but 
that would require a more extensive Table than the one given in 
Art. 334. 

(324.) Logarithms are attended with still greater advantages 
in the involution of powers, and in the extraction of roots. For 
if we raise both members of equation (1) to the nth power, we 


obtain 
N* = a”. 


~~ 


ye * LOGARITHMS. 


Therefore, according to the definition, mx is the logarithm of N’, 
since nz is the index of that power of the base which is equal to 
N”; that is to say, 

The logarithm of any power of a number, is equal to the logarithm 


/ of that number multiplied by the exponent of the power. 


EXAMPLES. 


Ex. 1. Find the third power of 4 by means of logarithms. 


The logarithm of 4 is 0.602060 
Multiply by 3 
The product is 1.806180, 


which is the logarithm of 64. 
Ex. 2. Find the fourth power of 3 by means of logarithms, 
Ex. 3. Find the seventh power of 2 by means of logarithms. 


Ex. 4. Find the third power of 5 by means of logarithms. | 
Again, if we extract the mth root of both members of equation 
(1), we shall obtain 


1 nd 
N? =a" 

x re 
therefore, according to the definition, 5 is the logarithm of N*; 


that is to say, 


The logarithm of any root of a number, is equal to the logarithm of 
that number divided by the index of the root. 


EXAMPLES. 


Ex. 1. Find the square root of 81 by means of logarithms. 


The logarithm of 81 is 1.908485 
Divided by 2 
The quotient is 954243, 


which is the logarithm of 9. 
Ex. 2. Find the square root of 121 by means of logarithms. 
Ex. 3. Find the sixth root of 64 by means of logarithms. 


Ex. 4. Find the third root of 125 by means of logarithms. 
The preceding examples will suffice to show, that if we had 


LOGARITHMS. 299 


tables which gave the logarithms of all numbers, they would 
prove highly useful when we have occasion to-perform frequent 
multiplications, divisions, involutions, and extraction of roots. 


(825.) The following examples will show the application of 
some of the preceding principles. 


Ex. 1. log. (abcd) = log. a+ log. b + log. c + log. d. 

Ex. 2. log. (=) = log. a + log. 6 + log. c— log. d— log. e. 

Ex. 3. log. (a"b"c”) = m log.a+n log.b+ p log. c. 

Ex. 4. log. € ~) = =m log.a+n log. b—p log. c. 

Ex. 5. log. (a? — a’) = log. [(a+ 2x) (a—2z)| = log. (a+ 2) + 
log. (a — 2). 

Ex. 6. log, Va? —2? = } log. (a+ 2) + 3 log. (a—z) 


eat 15 
Ex. 7: log. a* */ a’ = log. er 15 log. a. 


(326.) We shall presently explain a method by which logarithms 
may be computed. We may observe, however, that it is not 
necessary to compute the logarithms of all numbers independently. 
From the logarithms of a few numbers we may readily derive 
the logarithms of a great many other numbers. 

We have seen, in Art. 323, that the logarithm of a product 1s 
found by adding nyse the logarithms of the factors. Let us 
represent the apes of 2 by #; then, since the logarithm of 
10 is 1, we shall have 


o 


log.20 =2+1, log. 20000 = 
log. 200 =2+2, log. 200000 = 
~ log. 2000 = « + 3, log. 2000000 =, &c. 
We have seen, in Art. 324, that the logarithm of any power of 
a number is equal to the logarithm of that number multiplied by 
the exponent of the power. 
Hence, log. 4 == 2x, log. 32 = 
Og 2. saan log. 64 = 
log. 16 = 4a, log. 128 =, &c. 


300 


LOGARITHMS. 


Hence we find, also, that 


log. 40 =2%r+1, 
log. 400 = 2x + 2, 
log. 80 =32+41, 
log. 800 = 32 + 2, 
log. 160 =4a + 1, 
log. 1600 = 4a + 2, 


log. 
log. 
log. 
log. 
log. 
log. 


4000 = 
40000 =, &e. 
g000 = 
80000 =, &e. 
16000 = 
160000 =, &e. 


We have seen, in Art. 323, that the logarithm of a fraction 1s 
equal to the logarithm of the numerator minus the logarithm of 


the denominator. 


Hence,log.50 =2— a, 
log.500 =3— a, 
log. 25 =—2— Qa, 
log. 125 =3— 32a, 


log. 


log. 


.250 =3 — 2z, 
~ 2500 = 4 — 2z, 
gs. 1250 =4— 3a, 
. 12500 = 5 — 3z, 


6250. —5 — 4z, 
62500 — 6 — 4a, 


Hence, log. (4) = log. 5 = 1—za. 


. 50004 
. 50000 =, &e. 
. 625 = 
. 3125 =, oe. 
25000 ees 
. 250000 =, &ec. 
sa 2OOUU ee 
> 1250000 ES, occ: 
“P2000 te 
- 6250000 =, &c. 


(327.) So, also, from the logarithm of 3 we might easily derive 
a great number of other logarithms. 
308, we find the logarithm of 3 to be .477121; it is required to 
derive from this the logarithm of 30. | 


Required the logarithm of 3000. 


Required the logarithm of 9. 


Required the logarithm of 27. 
Required the logarithm of 81. 
Required the logarithm of 90. 
Required the logarithm of 270. 
Required the logarithm of 900. 


From the Table on page 


From the same Table, we find the logarithm of 2 to-be .301030. 
[t is required, by the aid of the logarithms of 3 and 2, to obtain 
the logarithm of 6. 


LOGARITHMS. . 301 


Required the logarithm of 12. 
Required the logarithm of 15. 
Required the logarithm of 18. 


From the same Table, we find the logarithm of 5 to be .69897U. 
It is required from this to deduce the logarithm of 50. 


Required the logarithm of 500. 
Required the logarithm of 5000. 


From the same Table we find the logarithm of 95 to be 1.977724. 
The logarithm of 9.5, or 25, is equal to the logarithm of 95 minus 
the logarithm of 10. 


Hence, the logarithm of 9.5 is 0.977724. 
Also, the logarithm of 950 is 2.977724. 


Hence the decimal part of the logarithm of any number is the same 
as that of the number multiplied or divided by 10, 100, 1000, &e 

Prime numbers are such as cannot be decomposed into factors, 
as, 2, 3, 5, 7, 11, 13, 17, &c. All other numbers arise from the 
multiplication of prime numbers. If, therefore, we knew the 
logarithms of all the prime numbers, we could find the logarithms 
of all other number by simple addition. 


(328.) We will now explain a method by which the logarithm 
of any number may be computed. 

If a series of numbers be taken in Geometrical progression, their 
logarithms will form a series in Arithmetical progression. Thus, 
take the geometrical series 


1, 10, 100, 1000, 10000, 100000, 


their logarithms are 
0, 1, 2, 3, 4, D5 


forming an arithmetical series. 

If, now, we interpolate a geometrical mean between any two 
numbers in the first series, its logarithm will be the arithmetical 
mean between the two corresponding numbers in the lower series. 

Find, for example, a geometrical mean between 1 and 10. It 
will be the square root of 10, or 3.162277. The arithmetical 
mean between 0 and 1 is 0.5. 

Therefore, the logarithm of 3.162277 is 0.5. 


302 LOGARITHMS. 


Find, again, a geometrical mean between 3.162277 and 1, 
which is 5.623413. Find, also, the arithmetical mean between 0.5 
and 1, which is 0.75. 

Therefore, the logarithm of 5.623413 is 0.75. 

Find now a geometrical mean between 3.162277 and 5.623413, 
which is 4.216964. Its logarithm will be the arithmetical mean 
between 0.5 and 0.75, which is 0.625. 

Therefore, the eatin of 4.216964 is 0.625. 

Find, again, a geometrical mean between 4.216964 and 5. 623415, 
which is 4. 869674. Its logarithm will be the arithmetical mean 
between 0.625 and 0.75, which is 0.6875. 

Thus we have found the logarithms of four new numbers, and 
in this manner we might proceed to construct a table of loga- 
rithms. It will be observed, however, that these numbers are all 
fractional, whereas it is most convenient to have the logarithms 
of integers. By pursuing this method, however, we might event- 
ually find the logarithm of a whole number ; as, for example, 5 
For we have already found the logarithm of 


5.623413 to be 0.75, 
and the logarithm of 4.869674 ‘ 0.6875. 


One of these numbers is greater than 5, and the other less. A 
geometrical mean between them is 5.232991, which is too great ; 
but the mean between this result and the last of the two preceding 
is 5.048065, which is already a close approximation. By pur- 
suing the same method, we may come nearer and nearer to the 
number 5, until at last, after finding twenty-two geometrical 
means, the difference is inappreciable in the sixth decimal place 
and we obtain 


the logarithm of 5 equal to 0.69897 ; 


and by a like process, the logarithm of any peaar number may be 
found. 


(329.) Hence, to compute the logarithm of any number, we 
have the following 


RULE. 


Take the geometrical series 1, 10, 100, 1000, 10000, &c., and apply 
to it the arithmetical series 0, 1, 2, 3, 4, &e., as logarithis. 


LOGARITHMS, 303 


Find a geometrical mean between 1 and 10, 10 and 100, or any other 
two terms of the first series between which the proposed number lies. 

Between the mean thus found and the nearest extreme, find another 
geometrical mean in the same manner, and so on, till you arrive within 
the proposed limit of the number whose logarithm is sought. 

Find, also, as many arithmetical means between the corresponding 
terms 0, 1, 2, 3, 4, &c., of the other series, in the same order as you 
found the geometrical ones ; the last of these will be the logarithm 
answering to the number required. 

In this manner were the logarithms of all the prime numbers at 
first computed ; but much more expeditious methods have since 
been devised. 

Having obtained the logarithm of 5, it is easy to find the loga- 
rithm of 2. For the logarithm of 2 = log: (49) = log. 10 — log. 
5 = 1— 0.69897 = 0.30103. 


(330.) We have seen, in Art. 322, that 


the logarithm of 0.1 is — 1, 
é¢ ee 0.01 “ —2, 
of f 0.001 “ — 8, 
i ff 0.0001 “ — 4, 
i vs * 0.00001 “* — 5, &e. 


That is, the smaller the fraction, the greater its logarithm 
Hence, if the fraction be infinitely small, its logarithm will be in- 
finitely great; that is, in Briggs’ system, the logarithm of 0 is in- 
finite and negative. 


» 


LOGARITHMIC SERIES. 


(331.) We have already explained a method of computing loga- 
rithms ; but this method is very laborious in practice. It is found 
much more convenient to express the logarithm of a number in 
the form of a serves. 

Let y be a number whose logarithm is required to be developed 
in a series, and let us employ the method of Unknown Coefficients. 
Put y = 1+ 2, and assume 


log. (1+ a) = Aw + Ba? + Cx* + Dat+,&e., (1). 


304 LOGARITHMS. 
Assume, iso, 
log. (1 + z) =“Az + B2’ + C2* + D2 +, &e. (2). 
Subtracting equation (2) from (1), we obtain 


log. (1+ 2) —log.(1+ z) =A(x— z) + B(a’ — 2’) + C(a#®— 2°) 4, 


&e. (8). “ 


The second member of this equation is divisible by r —z3; we 
will reduce the first member to a form in which it shall also be 
divisible. 


We have log. (1 + x) — log. (1 + 2) = log. (- it) = log. 


L—2 
(: ek add 
But, since eee may be regarded as a single quantity, v, we 


may develop log. (1+ v) in the same manner as log. (14 2), 
which gives 


L—2z r—z x—2z\* x—z* 
as (+753) =4 ieet B(Ge) +8 a) +180. 


This last series must be identical with the one which we have 





already obtained for log. (1 fe i): or its equal, log. (1 + 2) | 


— log. (1+ 2), in equation (3) ; and since the terms of both are 
divisible by « — z, by taking out this common factor, we obtain 


1 L—z (x—z)’ 
A.——+ B——, + C+ 
pet (pay “(dep 
rz + 2°)+, &e. 
Since this equation, like the preceding, must be verified for all 
values of x and z, the equality must subsist when a = z. But on 


this hypothesis, all the terms of the first series vanish except one, 
and we have simply 
as = A+ 2Ba + 3C2? + 4Dax° + 5Ea* +, &c.; 


or, performing the division indicated in the first member, 


A(l—«2+a°—2+a*—...) = A+ 2Ba + 30x’? + 4D2* 4+... 


3 +, &e., =A + Biv + 2) + O(@ + 


Spi 


LOGARITHMS. 305 


Therefore, according to the principle of Art. 287, we have the 
equations 


A=A, 
A 
—A = 2B; whence B=— >. 
A=30; « C=42 
A 
i, A. 6c a BS 
A= 4D D= a 
A 


A — 5E 3 ie E — +. 5° 

The law of the series is obvious; and hence, substituting the 
values of B, C, D, &c., in equation (1), we obtain, for the develop- 
ment of log. (1+ 2), 


log. (I+2)=>.2—Satee— pat... 
De a he ee Saeed ae 
= eres est gery fal) 


The number A is called the modulus of the system of logarithms 
employed. Lord Napier, the illustrious inventor of logarithms, 
assumed the modulus equal to unity. If, then, we designate Na- 
perian logarithms by log.’, we shall have 

|S ec a 
log. (l+2)=7 a tomer SL g + &e. (4). 

By giving to x in succession all possible values, we may obtain 

from this equation the logarithms of all numbers. 


If we make x — 0, we shall have log.’ 1 = 0. 
Make x = 1, and we obtain | 
log’ 2=1—1+41—1+41~—, &e, 
a series which converges so slowly that it would be necessary to 
employ a very large number of terms to obtain the accuracy de- 
sirable. The series may be rendered more converging in the 


following manner: 
In equation (4), substitute — x for 2, and it becomes 


log.’ (l—2) =—-~—5—3s-—7 ,&e. (5). 


306 _ LOGARITHMS. 


Subtracting equation (5) from equation (4), and observing that 





log.’ (1+ x) — log.’ (1— 2) = log.’ set 


(As is Fale ee Ae ‘ 
log. (ma)oGreta tits to). 


x . 
» we obtain 
= | 





aaa 


Put ——_ = 1+ — = (2 being at least eoual to unity), and we 
1 
have (1+ 2)z=(1—2) (z+ 1); whence, by reduction, z= ET 
Hence the preceding series becomes 
/ / — | 1 a 1 —". 
log.’ ou 5) =log: (2 + 1) —log. 2=2 (gory Homa 


CEN, M52 ..). 


(332.) The last series may be employed for computing the 
logarithm of any number, when the logarithm of the preceding 
number is known. Making successively z ='1, 2, 4, 6, &c., we 
find the following 


NAPERIAN, OR HYPERBOLIC LOGARITHMS. 

















log 2 = 22 + +e ee x = 0.693147 
log.’ 3=log.’2 ‘ Pa arte ea tee) = 1.098612 
log’ 4=2 log.’ 2 = 1.386294 
log.’ 5:5 Jog, rye, i H.. :) = 1.609438 
log./ Ee eedag ks ae BP 2 
log.’ 7 =log/ 6 + 2(4 ee oe oe ee .) = 1.945910 
log.’ 8 =log’ 4+ rat 2 = 2.079442 
logs 9: 2log 3 = 2.19 7eeo 
log.’ 10 = log.’ 5 + log.’ 2 = 2.302585 
&c., &c., &e. 


(333.) The Naperian logarithms being computed, it is easy to” 
form any other system. We have ie 


eo We. A a ae) ae 
log. (1-0) =A(F—F4G—T4+5—G te). 


LOGARITHMS. 307 


Distinguishing the Naperian logarithms by an accent, we have 
Gy cs ME a A GAY 
log. (1 +2) =a/(F ye Zt Ee G07 
Hence . 
log. (1+): log.’ (1+ a)::A:A’ 
Therefore, the logarithms of the same number in different systems 


are to each other as the moduli. 
In Napier’s system, the modulus = 1. Hence 


log. (1+27)=A.log’ (1+ 2). 

That is, the common logarithm of a number is equal to its Naperian 
logarithm multiplied by the modulus of the common system. 

If, then, we knew the modulus of the common system, we could 
easily convert the preceding Naperian logarithms into common 
logarithms. Now, from the equation 

log. (1+ 2)—A.log.’(1+ 2), we obtain 
__ log. (1+ 2) 
~ log’ (1+ a) 
log. 10 


Suppose z= 9, then A = Tog! 10" 
But log. 10 =1. Hence 


ieee tet 
— Tog.’ 10 — 2.302585 


which is the modulus of the common system. 


— 0.434294, 


(334.) We can now compute the 


COMMON, OR BRIGGS’ LOGARITHMS. 


log. 2 = 0.434294 x 0.693147 = 0.301030 
log. 3 = 0.434294 x 1.098612 = 0.477121 
log. 4= 2log. 2 = 0.602060 
log. 5= log. 10—log.2= 1— log. 2 = 0.698970 
log. 6=—log. 3+4 log.2 = 0.778151 
log. 7 = 0.434294 x 1.945910 — 0.845098 
log. 8= 3log.2 = 0.903090 
log. ees log. 3 = 0.954243 
log. 10 = = 1.000000 


Sc. orb &e. 


308 


LOGARITHMS. 


We thus obtain the following Table of Common Logarithms : 


— 





















































































L ogarithm. — 
106 | 2.025306 
107 | 2.029384 | 
108 | 2.033424 
109 | 2.037426 
110 | 2.041393 | 
111 | 2.045323 | 
112 | 2.049218 
113 | 2.053078 





114 | 2.056905 
115 | 2.060698 | 
116 | 2.064458 | 
2.068186 


118 | 2.071882 | 


119 | 2.075547 
120 | 2.079181 | 
121 | 2.082785 
122 | 2.086360 





123 | 2.089905 
124 | 2.093422 | 


125 | 2.096910 | 


126 | 2.100371 
127 | 2.103804 | 
128 | 2.107210 | 
129 | 2.110590 | 


130 | 2.113943 























131 | 2.117271 
2.120574 
133 | 2.123852 | 
134 | 2.127105 
135 | 2.130334 | 
. | 101 | 2.004321 | 136 | 2.133539 | 
137 see 
138 | 2.139879 
139 | 2.143015 


| No _ Logarithm. || No _ Logarithm. || No. | Logarithm. || No. 

~ 11 0.000000 | 36 | 1.556303 | 71 | 1.851258 
2 | 0.301030 | 37 | 1.568202 || 72 | 1.857333 
3 | 0.477121 | 38 | 1.579784} '73 | 1.863323 
4, | 0.602060 || 39 | 1.591065 || 74 | 1.869232 
5 | 0.698970 || 40 | 1.602060 | 75 | 1.875061 
6 | 0.778151 || 41 | 1.612784 || 76 | 1.880814 
7 | 0.845098 || 42 | 1.623249 || 77 # 1.886491 
8 | 0.903090 || 43 | 1.633468 || 78 | 1.892095 
9 | 0.954243 | 44 | 1.643453} 79 | 1.897627 
10 1.000000 | 45 | 1.653213 |) 80} 1.903090 
11 | 1.041393 | 46 | 1.662758 || 81 | 1.908485 

12 | 1.079181 | 47 | 1.672098 || 82 | 1.913814 || 117 
13 | 1.113943 || 48 | 1.681241] 83] 1.919078 
14. | 1.146128 || 49 | 1.690196 || 84] 1.924279 
15 | 1.176091 | 50 | 1.698970 || 85 } 1.929419 
16 | 1.204120 1.707570 || 86 | 1.934498 
17 | 1.230449 || 52! 1.716003 | 87] 1.939519 
18 | 1.255273 || 53 | 1.724276 | 88 | 1.944483 
19 | 1.278754 || 54 | 1.732394 || 89 | 1.949390 
20 | 1.301030 | 55. 1.740363 1.954.243 
91 | 1.322219 || 56 | 1.748188 || 91] 1.959041 
22 | 1.342423 | 57 | 1.755875 | 921 1.963788 
23 | 1.361728 || 58 | 1.763428 | 93 | 1.968483 
| 24 | 1.380211 | 59 | 1.770852 | 94 | 1.973128 
25 | 1.397940 || 60 | 1.778151) 95 | 1.977724 

26 | 1.414973 || 61 | 1.785330 || 96 | 1.982271 | 

27 1.431364 | 62 | 1.792392 | 97 | 1.986772 | 132 
28 | 1.447158 | 63 | 1.799341 | 98 | 1.991226 
29 1.462398 || 64 | 1.806180 || 99 | 1.995635 
30 | 1.477121 || 65 | 1.812913 | 100 | 2.000000 

31 | 1.491362 1.819544 || 101 | 2.004321 || 136 
32 1.505150 || 67 | 1.826075 || 102 | 2.008600 
oy 1.518514 || 68 | 1.832509 || 103 | 2.012837 

a 1.531479 || 69 | 1.838849 | 104 | 2.017033 | 

5 | 1.544068 | 70 | 1.845098 || 105 | 2.021189 

















140 | 2.146128 | 





(335.) Let us now determine the base of Napier’s eyateran 
Designating it-by a, we shall have, Art. 333, 


But log.a= 1. 


log.’ 


a:log.a:: 


Hence 


1 : 0.434294, 


log. a = 0.431294, 


* 


> * 
i» : 
- + 
- 
¢ 
- 


LOGARITHMS. 309 


That is, the modulus of the common system is eet to the common 
logarithm bf Napier’s base. 

We wish, then, to find the number corresponding to the com- 
mon pearithit 0.434294. By inspecting the preceding table, we 
see that this number must be a little less than 3. More accu- 
rately, it is 

2.718282, 


which is the base of Napier’s system. 

Any number, except unity, may be taken as the base of a sys- 
-tem of logarithms, and hence there may be an infinite number of 
systems. Only two systems, however, are much used; those of 


Briggs and Napier. 
The base of _ Briggs’ system is 10. 


Jp Napier’s fo 2.718282. 
The modulus of Briggs’ i 0.434.294, 
as Napier’s af 1. 


Hence, in Briggs’ system, all numbers are to be regarded as 
powers of 10 


Thus, LO i= 2} 
UV edt ee | 
Megan lpeem 22 
{AY pomfooe dD 
Or. OC. 


In Napier’s system, all numbers are to be regarded as powers 
of 2.718282. | 


Thus, 2.7189 = 2, 
DARL SE ee 53. 
2. tLe em 4, 
Bei 13'S = fh 
Ren ec: 


Briggs’ logarithms are employed in all the common operations 
of multiplication and division, and hence they are known by the 


310 LOGARITHMS. 


name of common logarithms. Napier’s logarithms are of great 
use in the application of the calculus to many analytical and 
physical problems. They are also called Ayperbolic logarithms, _ 
having been originally derived from the hyperbola. 


EXPONENTIAL EQUATIONS. 


(336.) An exponential quantity'is one which is raised to some 
unknown power, or which has an unknown quantity for an og 
nent; as, 

3 £ 
OM Pe", OF | REC. 

An exponential equation is one which contains an exponential 
quantity ; as, 

Sf ee b, = a, &e. 

Such equations are most easily solved by means of logarithms. 
Thus, consider the equation 


a’ = b. 
Taking the logarithm of each member of the equation, we have 
x log. a = log. 8, 
log. 6 
log. a 


Ex. 1. What is the value of x in the equation 3° = 811 


By the preceding formula, = 5 ue 





OY Shue! iro 


Looking out the logarithms of 81 and 3 from the Table i in Art. 
334, we have 


_ 1.908485 _ 
ATES 


Therefore, 3‘ — 81. 
Ex. 2. What is the value of x in the equation 3” = 202 


_ log. 20 1.301030 
ae ty Ra 7 101 


Therefore, 3°77 — 20 nearly. 


= 2.727 nearly. 


LOGARITHMS. ape 


(337.) The other equation, z” = a may be solved by ¢rial, as in 
Art. 319. Thus taking the logarithm of each member, we have 


xv log. x = log. a. 


Find now, by trial, two numbers nearly equa: to the value of z, 
and substitute them in the given equation instead of the unknown 
quantity. Then say, 


As the difference of these results, 

Is to the difference of the two assumed numbers, 

So zs the error of either result, 

To the correction required in the corresponding assumed number. 


Ex. 1. Given «* = 100 to find the value of z. 

Here we have x log.x—log. 100 = 2. 

Suppose fits 

Then 0.477121 x 3 = 1.431363, which is too small. 
Suppose t= 4, 

Then 0.602060 x 4 = 2.408240, which is too great. 


Hence the value of x is between 3 and 4, but nearer to 4. 
Assume, then, 3.5 and 3.6 for the two numbers. 


By the first supposition, By the second supposition, 

2=3.5; log.r— .544068 @ = 3.6; log. 2 =, 56303 

Multiplied by 3.5 Multiplied by 3.6 
z ‘log. « = 1.904238 x. log. 2 = 2.002689 
Diff. of results : Diff. assumed numbers : : Error of 2d result : Its per ection. 

098451: 0.1 os 002689 =: .00273 
Hence 2 = 3.6 — .00273 = 3.59727 nearly. 
Therefore, S5972T "=" — 100mearly, 


{If we wish a more accurate result, the operation must be re- 
peated with two new numbers; as, for example, 3.59727 and 
3.59728. 


Ex. 2. Given z” = 202, to find an approximate value of z. 


312 LOGARITHMS. 


COMPOUND INTEREST. 


(338.) In calculating compound interest, the first subject of 
inquiry is, to what sum does a given principal amount, after a 
certain number of years, the interest being annually added to the 
principal? It is evident that $1.00, placed out at 5 per cent., be- 
comes at the end of a year, a principal of $1.05. But the amount 
at the end of each year must be proportioned to the principal at 
the beginning of the year. In order, then, to find the amount at 
the end of two years, we say 

LOO Ze OD ei los Lule 
To find the amount at the end of three years, we say 
LOO’ LOD t): (105) 211,00). 

And in the same manner we find that the amount of $1.00 for 
m years at compound interest is (1.05)”. 

If the rate of interest were six per cent., we should find the 
amount for m years to be (1.06)”. 

The amount of two dollars for a given time must obviously be 
double the amount of one dollar, and the amount of $1000 must 
be a thousand times the amount of one dollar. 

Hence, if we put P to represent the principal, 

r the rate per cent. considered as a decimal, 
nm the number of years, 
A the amount of the given principal for 2 years, 
we shall have 
A=P. (1 -+- Bye 

This equation contains four quantities, A, P, m, 7; any three 
of which being given, the other may be found. The computations 
are most readily performed by means of logarithms. Taking the 


logarithms of both members of the preceding equation and re- 
ducing, we find 


1. Log. A =n xX log.(1+ 7r) + log. P. 
2. Log. P = log. A—n x log.(1+ 7). 
8. Log. (1 +7) = chem she 

log. A — log. P 


4s, 2 — 





J log. (1 ae r) 


LOGARITHMS. 313 


EXAMPLES. 


Ex. 1. What is the amount of twenty dollars, at 6 per cent. 
' compound interest, for 11 years % 
Here we employ formula (1). 


Amount of $1.00 for 1 year $1.06, log. = 0.025306 


Multiplying by 11, 11 
0.278366 

Given principal $20. log. = 1.301030 
Amount $38 nearly, 1.579396 


This result is derived from the Table on page 308. By con- 
sulting a larger Table, we should find the amount $37.966. 

Ex. 2. What principal at 5 per cent. interest will amount to $50 
in 13 years? 

Here we employ formula (2). 


1+7r= 1.05, log. = 0.021189 
Multiplying by x, 13 
Subtract 0.275457 
From log. A b = 1.698970 
P = $26.5 nearly, ~ 1.423513 


More accurately, P = $26.516. 

Ex. 3. At what rate per cent. must $40 be put out at compound 
interest, that it may amount to $57 in 9 years? 

Here we employ formula (3). 


A =5i7, log. = 1.755875 

P — 40, log. = 1.602060 

Dividing by 2, 9)0.153815 
ee 104” — 0.017091 


Consequently,r = .04, or four per cent. 
Ex. 4. In what time will $50 amount to $90 at 5 per cent. 
Here we employ formula (4). 
AS 90, log. = 1.954243 
P= 50, log. — 1.698970 


‘1+7r= 1.05, whose logarithm is 0.021189)0.255273 
Dividing one logarithm by the other, we obtain 12, Ans. 


# 


314 LOGARITHMS. 


Ex. 5. What is the amount of $52 at 3 per cent. compound in- 
terest for 15 years? 
Ans. $81. 
Ex. 6. What principal at 6 per cent. compound interest will 
amount to $101 in 4 years? 
Ans. $80. 
Ex. 7. At what rate will $10 amount to $16 in 16 years? 
Ans. Three per cent. 
Ex. 8. What will $300 amount to in 10 years at compound in- 
terest semi-annually, the yearly rate being 6 per cent. ? 
Ex. 9. In what time will a sum of money double at 6 per cent. 
compound interest ? 
Ans. 11.89 years. 
Ex. 10. In what time will a sum of money ariple itself at 4 per 
cent. compound interest 1 
Ans. 28.01 years. 
(339.) The natural increase of population in a country may be 
computed in the same way as compound interest. Knowing the 
population at two different dates, we compute the rate of increase 
by formula (3), and from this we may compute the population at 
any future time on the supposition of a uniform rate of increase. 


EXAMPLES. 


Ex. 1. The number of the inhabitants of the United States in 
1790 was 3,900,000; and in 1840, 17,000,000. What was the 
average increase for every ten years? 

Ans. 34 per cent. 

Ex. 2. Suppose the rate of increase to remain the same for the 
next ten years, what would be the number of inhabitants in 18501 

Ans. 22,800,000. 
Ex. 8. At the same rate, in what time would the number in 
1840 be doubled 2 
Ans. 23.54 years. 
Ex. 4. At the same rate, what was the population in 17801 
Ans. 2,900,000. 

Ex. 5. At the same rate, in what time would the number in 
1840 be tripled? 

Ans. 37.31 years. 
THE END. 


HARPER & BROTHERS, NEW YORK, 


HAVE RECENTLY PUBLISHED 


A Creatise on Algebra, 


BY ELIAS LOOMIS, 


PROFESSOR OF MATHEMATICS AND NATURAL’ PHILOSOPHY IN THE UNIVERSITY OF THE 
CITY OF NEW YORK. 


‘ 


1 volume octavo, Sheep. $1 25. 


This Treatise was specially designed for the use of the Students of the New 
York University, but it is believed to be adapted to the wants of Students gen- 
erally in the Colleges and Academies of the United States. The following 
notices are deemed sufficient to call the attention of teachers to an examina- 


tion of this work. 


I have carefully examined the work of Professor Loomis on Algebra, and am much 
pleased with it. The arrangement is sufficiently scientific, yet the order of the topics 
‘is obviously, and, I think, judiciously made with reference to the development of the 
powers of the pupil. The most rigorous modes of reasoning are designedly avoided 
in the earlier portions of the work, and deferred till the student is better fitted to ap- 
preciate them. All the principles are, however, established with sufficient rigor to 
give satisfaction. Much care seems to have been taken, by generalizing particular 
examples and other means, to develop the faculty and induce the habit of general- 
izing, a point which, I think, has not received sufficient attention hitherto. On the 
whole, therefore, I think this work better suited for the purposes of a text-book than 
any other I have seen.—AuGustUs W. Smiru, A.M., Professor of Mathematics and 
Astronomy in the Wesleyan University. 


Professor Loomis’ Treatise on Algebra is an excellent elementary work. It is suffi- 
ciently extensive for ordinary purposes, and is characterized throughout by a happy 
combination of brevity and clearness.—ALEXxIs CaswELL, D.D., Professor of Mathe- 
matics and Natural Philosophy in Brown University. 


I have examined Prof. Loomis’ new work on Algebra, and am highly pleased with 
it. For conciseness and clearness of statement, and for its lucid explanation of ele. 
mentary principles, it is decidedly superior to any work with which I am acquainted. 
I hope it will be extensively used in all our public institutions ——ALoNnzo Gray, A.M., 
Professor in Brooklyn Female Academy. 


I have examined Prof. Loomis’ Algebra carefully and with much interest, and am 
so perfectly satisfied with it, that I shall introduce it to my classes as soon as possi- 
ble. It is just the work which I have been for a long time in search of. I am par- 
ticularly delighted with the mode of treating the subject of logarithms, and, indeed, 
with the clearness of the investigations generally throughout the work.—H. Oris 
KeEnpatt, Professor of Mathematics and Astronomy in the Central High School of 
Philadelphia. 


I fully concur with Prof. Kendall in his opinion of Loomis’ Algebra.—Srars C. 
Watker, of the National Observatory, Washington. 


A text-book like this of Prof. Loomis’ was much needed, and the desideratum is 
so well supplied, that I think it can not fail to commend itself at once to the favorable 
regard of those who are looking for the best work for college classes. I consider it 
decidedly the best book for college instruction that I am acquainted with on the sub- 
ject, and it has been adopted as a text-book in our college by unanimous consent of 
the faculty. Prof. Loomis has been very happy in simplifying the more difficult parts 
of the subject, especially on the theory of equations and on logarithms.—JAmEs Noon- 
EY, A.M., Professor of Mathematics and Natural Philosophy in Western Reserve 
College. 


I have carefully examined Prof. Loomis’ Algebra, and think it better adapted for 
a text-book for college students than any other I have seen.—C. GILL, Professor of 
Mathematics in St. Paul's College. 


2 Critical Notices of Loomis’ Algebra. 


Prof. Loomis seems very happily to have observed the proper medium between 
exuberance of explanation and demonstration on the one hand, which leaves little or 
nothing for the student himself to do; and a repulsive conciseness on the other, which 
discourages him, and gives him a disrelish for this portion of study. Ihave adopted 
it as a text-book in the Cornelius Institute, believing it to be better suited to youth 
who are preparing for college, than any other treatise on Algebra with which I am 
acquainted.—JoHNn J. OWEN, A.M., Principal of the Cornelius Institute, N. Y. City. 


Prof. Loomis’ work on Algebra is exceedingly well adapted for the purposes of in- 
struction. He has avoided the difficulties which result from too great conciseness, and 
aiming at the utmost rigor of demonstration; and, at the same time, has furnished 
in his book a good and sufficient preparation for the subsequent parts of the mathe- 
matical course. I do not know of a treatise which, all things considered, keeps both 
these objects so steadily in view. I regard the methods of explanation and illustra- 
tion as quite happy and satisfactory, and hope the work may be extensively used.— 
J. Warp ANDREWS, A.M., Professor of Mathematics and Natural Philosophy in Ma- 
rietta College. 


Prof. Loomis’ work is well calculated to impart a clear and correct knowledge of 
the principles of Algebra. The rules are concise, yet sufficiently comprehensive, 
containing in few words all that is necessary, and nothing more; the absence of which 
quality mars many a scientific treatise. The collection of problems is peculiarly rich, 
adapted to impress the most important principles upon the youthful mind, and the 
student is led gradually and intelligently into the more interesting and higher de- 
partments of the science. The theory of Permutations is happily introduced in con- 
nection with the involution of Binomials, and the chapter on Logarithms can not but 
be acceptable—Joun Brocxuessy, A.M., Professor of Mathematics and Natural 
Philosophy in Trinity College. 


I am much pleased with Prof. Loomis’ Algebra. I think he has accomplished very 
happily the object he had in view, and has prepared a work remarkably well adapted 
for the use of college students.—EBENEZER S. SNELL, A.M., Professor of Mathematics 
and Natural Philosophy in Amherst College. 


After a thorough examination of Prof. Loomis’ work on Algebra, I have concluded 
to adopt it as a text-book in this Institution—Marcus Catxin, A.M., Professor of 
Mathematics and Astronomy in Hamilton College. 


I am much pleased with Prof. Loomis’ Algebra. The arrangement of the subjects 
is, I think, an admirable one. The best proof I can give of the estimation in which 
I hold it, is the fact that at the proper time I intend to advise its adoption as the 
text-book on that subject in this college—Joun TatLock, A.M., Professor of Mathe- 
matics in Williams College. 


I have examined Prof. Loomis’ Algebra with great attention, and am so well pleased 
with its arrangement and execution throughout, that [ have decided to adopt as a text- 
book in this Institution—Joun E. SupLer, A.M., Professor of Mathematics in Dick- 
inson College. : 


Prof. Loomis has here aimed at exhibiting the first principles of Algebra in a form 
which, while level with the capacities of ordinary students and the present state of 
the science, is fitted to elicit that degree of effort which educational purposes require. 
Throughout the work, whenever it can be done with advantage, the practice is fol- 
lowed of generalizing particular examples, or of extending a question proposed rela- 
tive to a particular quantity, to the class of quantities to which it belongs; a practice 
of obvious utility, as accustoming the student to pass from the particular to the gen- 
eral, and as fitted to impress a main distinction between the literal and numeral cal- 
culus. The general doctrine of Equations is expounded with clearness, and, we may 
add, with independence. The author has developed this subject in an order of his 
own, Theorems which find a place in other treatises are omitted, and what some- 
times appears in a generic form, or in that of a corollary, becomes specific, or assumes 
the place of a primary proposition. We venture to say that there will be but one 
opinion respecting the general character of the exposition American Journal of Sci- 
ence and Arts. 


le a ee A ee 


WEBSTER’S OCTAVO DICTIONARY, REVISED, 


EMBRACING ALL THE WORDS IN THE QUARTO EDITION, AND 
ALSO AN ARRANGEMENT OF SYNONYMS UNDER THE 
LEADING WORDS. 


HARPER & BROTHERS, NEW YORK, 


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In one handsome Volume, of nearly 1400 pages, Sheep extra, Price $3 50, 


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THOROUGHLY REVISED AND CONSIDERABLY ENLARGED, 
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The most complete and thorough manual of 
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not fail to come into universal use, not only 
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ary World. 

This edition of Webster is all that could be 
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ceded it, and is a monument of learning and 
research.—NV. Y Commercial Advertiser. 

We can safely say that, for a dictionary for 
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The best English dictionary extant. Many 
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The whole work has been thoroughly re- 
vised by Prof. Goodrich, of Yale College, and 
several important and most valuable improve- 
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The labors of Professor Gooodrich have ma- 
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He has been engaged in them for three years 
past, and the application of his acute philo- 
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without ample fruit.—N. Y. Evening Post. 

It must be the standard English dictionary 
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public.—N. Y. Courier and Enquirer. 

Undoubtedly the best English dictionary 
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and in every possible way the work has been 
adapted to the wants of the great body of the 
people. It will find its way not only into all 
the schools and academies of the country, but 
to the desk of every student and the fireside 
of every family.—Mirror. 


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